Solve the following equation for x. Mathematics for the first year of junior high school .

1）a²(x-2)-3a=x+1

Solution: a x-2a -3a = x+1

a²-1)x=2a²+3a+1

When a is not equal to plus or minus 1, x=(2a +3a+1) (a -1) When a = +1, the equation has no solution.

When a=-1, x=arbitrary value.

2) ax+b-(3x+2ab) 3=1 2 solution: ax+b-x-2ab 3=1 2

a-1)x=1/2+2ab/3 - b

When a is not equal to 1, x = (3 + 4ab-6b) (6 (a-1)) when a = 1 and b = 3 10, x = any value.

When a = 1 and b is not equal to 3 10, the equation does not hold.

3）(x-b)/a=2-(x-a)/b

Solution: x a-b a=2-x b+a b

1/a+1/b)x=2+a/b+b/a

a+b)x ab=(a+b +2ab) ab knows that a and b are not 0 from the meaning of the title, and it is approximated.

a+b)x=(a+b)²

When a+b is not equal to 0, x=a+b

When a+b=0, x=arbitrary value.

1, (b-a)x=8, b=a, there is no solution.

When b is not equal to a, x=8 (b-a).

2, 1+m(x+1)=2m+1, mx+m=2m, mx=m, m=0, x is any number.

When m is not equal to 0, x=1

3, (1-m n)x = m+n = (n-m)x n, n is not 0

When n=m, m+n=2n is not 0 and has no solution.

When n is not equal to m, x = n(m+n) (n-m)

4, |x+2| +2x-3|= 5,x<-2, 2x<-4,x+2<0, 2x-3<-7<0,5=-x-2-2x+3=1-3x, x=-4 3>-2Contradiction.

2<=x<3 2, 2x<3, x+2>=0, 2x-3<0

5=x+2-2x+3=5-x, x=0.

3 2<=x, 3<=2x, 2x-3>=0, x+2>0,5=x+2+2x-3=3x-1, x=2

Combined, there is, x=0 or x=2

1. (b-a)x=8 So x=8 (b-a)(b≠a), b=a, there is no solution.

mx+m=2m+1, so, mx=m, x=13, x-mx n=m+n, x(1-m n)=m+n, so x=(mn+n) (n-m).

4、|x+2|+|2x-3|=5, order |x+2|、|2x-3|They are equal to 0, so x=-2, x=3 2

So when x -2, the equation is -x-2-2x+3=5, so when x=2x 3 2, the equation is x+2+2x-3=5, so when x=2-2 x 3 2, the equation is x+2-2x+3=5, so, x=0

(a-b)x=-8

x=-8/(a-b)

It is not equal to 0)m(x+1)=2(m+1 2)-1

x+1=(2（m+1/2)-1)/m

x=(2（m+1/2)-1)/m -1

1-m/n)x=m+n

x=(m+n)n/(n-m)

Let's simplify it again.

Absolute value + absolute value of 2x-3 = 5

Discuss on a case-by-case basis.

1.According to the conditions, x=8 (b-a), b≠a and 2x-3 0 with absolute values of 5, x=3 4x+2 0 and 2x-3 0, and the absolute values are 5, x=-3 4x+2 0 and 2x-3 0, and the absolute values are 5,x=0x+2 0 and 2x-3 0, and the absolute values are 5,x=-10

ax+3=bx-5

ax=bx-8

bx-ax=8

b-a)x=8

x=(b-a)/8

1+m(x+1)=2(m+1/2)

1+mx+m=2m+01

mx=m-1

x=1-1/m

x-m*x/n=m+n

xn-mx=nm+n*n

n-m)x=nm+n*n

x=(nm+n*n)/(n-m)

n is not equal to m because the divisor cannot be 0).

or the absolute value of -2x+2 + the absolute value of 2x-3 = 5

Absolute value of x + absolute value of 2x = 6

The absolute value of 3x = 6

x = 2 or -2

Solution (1): Shift the term to obtain: ax-4x=8+b

Merge to obtain: (a-4)x=8+b

The coefficient is one: x=(8+b) (a-4)(2) and the shift term is mx-nx=1

Solution: x=1 (m-n).

3) Remove the denominator to get: 4m(x-n) =3(x+2m) and remove the parentheses to get: 4mx-4mn=3x+6m

Shift: (4m-3)x=6m+4mn

The coefficient is one: x=(6m+4mn) (4m-3).

4x+b=ax-8

4-a）x=-8-b

a-4）x=8+b

x=a-8/4+b

mx-1=nx

mx-nx=1

m-n)x=1

x=1 (m-n) is written with parentheses removed.

1/3（x-n）=1/4(x+2m)

3（x-n）=4(x+2m)

3x-3n=4x+8m

x=-3n-8m

Then that's all I'll answer.

3）1/3m（x-n）=1/4(x+2m)4m(x-n) =3(x+2m)

mx-4mn = 3x+6m

mx-3x =6m+4mn

x =（6m+4mn）/ m-3x)

Solution: Let it start with x people, and then (x+5) people.

1 person 1 hour to complete 1 80

So x 2x1 80 + (x+5) x8x1 80 = 3 4 times 80 on both sides

2x+8(x+5)=60

2x+8x+40=60

10x=20

x=2x+5=7

A: It started with 2 people, and later it increased to 7 people.

Let's start with x people, and then x+5 people.

1 person 1 hour to complete 1 80

So x*2*1 80+(x+5)*8*1 80=3 4 multiplied by 80 on both sides

2x+8(x+5)=60

2x+8x+40=60

10x=20

x=2x+5=7

So it started with 2 people, and later it increased to 7 people.

Set up x people, known to be 1 80 efficacy

x/80*2+(x+5)/80*8=3/4x=2

Let a batch of data be 1, one person's efficiency is 1 80, and some people are set to x, so list the equation [(1 80)*x*2+(1 80)*(x+5)*8=3 4] to solve the equation, and we can get x as 2, where * is the multiplication sign. But I don't know if this will be suitable for the content of the first year of junior high school.

1*80h=80

Let's arrange x people first, that is, x*2h+(x+5)8h=80*3 4, and solve x=2 people.

Working speed d = 1 80

Suppose the number of some people is x, then.

2x/80+40/80=3/4

x=20 so first 20 people do 2h, then add 5 people to do 8h, and complete the work of 3 4

Idea analysis (for most radical equations):

Place the part with the root sign on one side of the equal sign, and the part without the root sign on the other side, both sides squared at the same time.

Solve equation solution: (x 2-2) = 14-7x .........1.

x^2-2=（14-7x）²…2.

x^2-2=196-196x+49x²……3.

Simplification yields 48x 2-196x+198=0

The solution yields x1 = 11 6 and x2 = 9 4

Inspection: 14-7x 0

x≤2x=9/4

x2=9 4 does not fit the topic, discard.

In summary: x=11 6

Solution: Deformation of the original equation:

x²-2) = 14 - 7x

Square on both sides: x -2 = (14-7x).

x²-2 = 14²-2x14x7x+(7x)²x²-2 = 196-196x+49x²

Move merge: 48x -196x + 198 = 0

24x²-98x+99 = 0

According to the root finding formula, x=[-b (b -4ac)] 2a , we get:

x=[98±√(98²-4x24x99)] / 48x=[98±√(9604-9504)] / 48=(98±10) 48

Therefore: x=9 4 or x = 11 6

Bring in the original equation for verification:

x=9 4 is its root, and x=11 6 is also its root, so the original equation has two roots:

x=9 4 and x=11 6

Solution: (x -2)+7x=14

x²-2）=14-7x

Both sides are squared.

x²-2=49x²-196x+196

48x²-196x+198=0

24x²-98x+99=0

6x-11)(4x-9)=0

x = 11 6 or x = 9 4

14-7x≥0

x 2x=9 4 is not satisfied, discarded.

x=11/6

√(x²-2)=14-7x

Square on both sides. x²-2=196-196x+49x²

48x²-196x+198=0

24x²-98x+99=0

4x-9)(6x-11)=0

x=9/4,x=11/6

After testing, x=9 4 substitution is not valid.

So x=11 6

√（x²-2) +7x=14

x -2) = 14-7x [x 2 and x 2] x -2 = (14-7x).

x²-2=196+49x²-196x

48x²-196x+198=0

4x-9)(6x-11）=0

x1=9/4,x2=11/6

Because x1>2 is not in harmony, x2 < 2, and x2 is greater than 2

Therefore, x2=11 6 is in line with the topic.

So x=11 6

Write it like you. It should be beyond the knowledge of junior high school. There should still be a root number after the square. It is recommended to send ** to follow-up questions.

If 7x is inside the root number, then x 9, 4, and 11, 6

The square of both sides is simplified to 50x 2=198 If you don't understand, you can ask questions.

mx-3=4x+n

mx-4x=3+n

m-4）x=3+n

x=（3+n）/(m-4)(m≠4)

ax-1=bx

ax-bx=1

a-b)x=1

x=1/（a-b）

If you don't understand, you can ask.

Remember, thank you if it helps.

1. (x m) 2+(1 n-n)x-1=0 (m≠0 and n≠0)=(1 n-n) 2+4 m 2>0 is constant, so there are two different real roots for x in the original equation.

x=/(2/m^2)=m^2/(2n)

2. The direct shift term has 2mn x=m 2-2nm+1 obviously m≠0 and n≠0, otherwise the above equation is not valid.

Therefore, the direct deformation has x=2mn (m 2-2nm+1).