Ask a few math problems with a high score 5x 1 3y 3 5x 3 4y 2 . Solve the equation 2x x 1 x 3x 2 x x

If (x 2+mx+8) (x 2-3x+n) does not contain x 3 and x 2 terms in the result, find the values of m and n:

x^2+mx+8)(x^2-3x+n)

x^4+(m-3)x^3+(8-3m+n)x^2+(mn-24)x+8n

Because (x 2+mx+8) (x 2-3x+n) does not contain x 2 and x 3 items.

So m-3=0

8-3m+n=0

m=3 n=1

m+n=3+1=4

Solve equation (x+3)(x+5)-(x-3)(x-5)=16:

x+3)(x+5)-(x-3)(x-5)=16

x²+8x+15-x²+8x-15=16

16x=16

x=1 solution equation: 2x(x-1)-x(3x+2)=-x(x+2)-12

2x(x-1)-x(3x+2)=-x(x+2)-12

2x^2-2x-3x^2-2x=-x^2-2x-12

2x^2-3x^2+x^2-2x-2x+2x=-12

2x+12=0

2(x-6)=0

x-6=0x=6 looks.

x+3)(x+5)-(x-3)(x-5)=16x^2+8x+15)-(x^2-8x+15)=1616x=16

x=1x^2+mx+8)(x^2-3x+n)x^4-3x^3+nx^2+mx^3-3mx^2+mnx+8x^2-24x+8n

x^4+(m-3)^3+(n-3m+8)x^2+(mn-24)x+8n

Because there are no x 3 and x 2 terms, that is, m-3=0 n-3m+8=0 solves m=3 n=1

1.Mountains and rivers, everywhere is clear and beautiful, sunny and rainy, always curious and curious 2Mountains and rivers are everywhere, bright and beautiful, sunny and rainy, so curious 3

Mountains and rivers, water everywhere, clear and beautiful, sunny and rainy, rainy, so curious 4Mountains and rivers, everywhere bright, bright, show! Rain or shine, always good, curious, strange!

5.Mountains and rivers, everywhere bright, bright, show! Rain or shine, always good, curious, strange!

6.Mountains and rivers, everywhere bright, bright, show! Rain or shine, always good, curious, strange!

7.The landscape is bright and beautiful, and it is curious when it rains. (Skipping words simplifies the reading method to make this couplet more concise and clear).

x+3)(x+5)-(x-3)(x-5)=16 His answer is x=1

For finding the method without x 3: multiplying x 2 with -3x in the first parenthesis will give x 3 a coefficient of -3, and multiplying mx with x 2 will give x 3 a coefficient of m, so that m-3 = 0 gives m=3

For finding without x 2: the method is the same as above, and the final n-3m+8=0 gives n=1

x+3)(x+5)-(x-3)(x-5)=16 solution: x 2+8x+15-(x 2-8x+15)=16, and finally x=1

x 2+mx+8)(x 2-3x+n) with x 3 is: -3x 3+mx 3 solution m=3

The term containing x 2 is: 8x 2+nx 2-3mx Substituting m=3 into the solution gives n=1

(1) 1/4x+2/5=1/2

Solution: Remove the denominator (multiply both sides of the equation by 20 at the same time) to get:

5x+8=10

Shift: 5x=10-8

Merge similar items to:

5x=2 coefficient to 1 yields:

x=；(2)x+x/7=3/14

Solution: Remove the denominator (multiply both sides of the equation by 14 at the same time) to get:

14x+2x=3

Merge similar items to:

16x=3 coefficient to 1 to obtain:

x=3/16。

3)13—5/3x=7x5/7+x

Solution: remove the denominator (multiply both sides of the equation by 3 at the same time) to get:

39-5x=15+3x

Shift: -5x-3x=15-39

Merge similar items to:

8x=-24

coefficient to 1 yields:

x=3。

Solution: Denominator: 18x+15=12x-2

Shift: 18x-12x=-2-15

Merge similar terms: (18-12) x = -17

Deparentheses: 6x=-17

Coefficient to one: x=-1 how can it be so complicated!!

Move items directly. The first question comes out...

The second question also moves the item directly...

What grade are you in !! Didn't the teacher hand in such a simple question?。。。

It's okay to move left and right, it's a matter of hurting you to help you, such a simple question.

Solution: 3x+5 2=2x-1 3 Left and right multiplication 6 to remove the denominator shift, merge the same terms, and reduce the coefficient to one.

（（1）2/5-3x=3/3-5x

Multiply both sides by 615-9x = 6-10x

10x-9x=6-15

x=-92)4/x+2-6/2x-3=1;

Multiply both sides by 123x+6-4x+6=12

4x-3x=6+6-12

x=0（3）12/x-1=15/2x

Multiply both sides by 605x-5=8x

8x-5x=-5

3x=-5x=-5/3

4）5/y+17-4/3y-7=-2

Multiply both sides by 204 (y+17)-5(3y-7)=-40

4y+68-15y+35=-40

11y=143

y=143÷11

y=13

8x+3y+2=0 ①

6x+5y+7=0 ②

5 gives 40x +15y +10 =0

3 gives 18x + 15y +21 =0 and subtracts from the equation: 22x - 11 =0

Solution x = 1 2

Substitution. 4+3y+2=0

The solution gives y = -2

So the solution of the system of equations is.

x= 1/2

y = -2

2x/3+3y/4=1/5

Multiply both sides by 6040x+45y=12 (1).

5x/6-5y/2=2

Multiply both sides by 65x-15y = 12 (2).

40x+15x=12+36

55x=48

So x=48 55

y=(5x-12)/15=-28/55

Write the wrong question, how can the first equation only have an x, what about y?

2x/3+3y/4=1/5...1) Multiply by 3 2 to get x+9y 8=3 10 move the term to get x=3 10-9y 8 Substitute x into (2) to solve it.

5x/6-5y/2=2...2)

x=48/55，y= —28/55