`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`

a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

b) \(\Rightarrow x^2-13x+22=0\)

\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)

c) \(\Rightarrow x^2-3x-10=0\)

\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, (x+2)2+(x-3)2=2x(x+7)

x.2+2.2+x.2+(-3).2-2x=8

2x+4+2x-6-2x=8

(2x+2x)+(4-6)=8

4x-2=8

4x=8+2

4x=10

   X=10:4

    X=5/2

a) (x+2)²+(x-3)² = 2x(x+7)

⇒x²+4x+4+x²-6x+9=2x²+14x

⇒x²+4x+4+x²-6x+9-2x²-14x=0

⇒ -16x+13=0

⇒ x=\(\dfrac{13}{16}\)

b) (x+3)(x²-3x+9) = x(x²+4)-1

⇒x(x²-3x+9)+3(x²-3x+9)=x³+4x-1

⇒x³-3x²+9x+3x²-9x+27-x³-4x+1=0

⇒-4x+28=0

⇒x=7

a: Ta có: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{x^2-4y^2}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(\)\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)

b: \(\frac{4x+7}{2x+2}-\frac{3x+6}{2x+2}\)

\(=\frac{4x+7-3x-6}{2x+2}\)

\(=\frac{x+1}{2\left(x+1\right)}=\frac12\)

c: \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)

\(=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\cdot\left(x+3\right)}\)

\(=\frac{x\left(x+9\right)-3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)

d: \(\frac{1}{x^2+3x+2}-\frac{1}{x^2-4}\)

\(=\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{1}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-2-\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{-3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)

a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.

b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .

a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .  

b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

giải phần còn lại giúp mình được ko?

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10

A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10

A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10

A = 0 + (3\(x-3x\)) - 10

A = 0  - 10

A = - 10 

dài thế ai trả lời đc hả ?

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