e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b)(x+3)²-(x-4)(x+8)=1

\(\Rightarrow\)x²+6x+9-(x²+8x-4x-32)=1

⇒x²+6x+9-x²-8x+4x+32=1

⇒2x+41=1

\(\Rightarrow\)2x+41-1=0

\(\Rightarrow\)2x+40=0

⇒2x=-40

\(\Rightarrow\)x=\(\dfrac{-40}{2}\)

⇒x=-20

\(x^2-3x+2.\left(x-3\right)=0\)

\(x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\left(x-3\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(x.\left(x-3\right)-3x+9=0\)

\(x.\left(x-3\right)-3.\left(x-3\right)=0\)

\(\left(x-3\right)^2=0=>x=3\)

a,\(x^2-3x+2\left(x-3\right)=0.\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

\(a)\frac{5}{8}-x=2\frac{1}{6}\)

\(\Rightarrow\frac{5}{8}-x=\frac{13}{6}\)

\(\Rightarrow x=\frac{5}{8}-\frac{13}{6}\)

\(\Rightarrow x=\frac{15}{24}-\frac{52}{24}\)

\(\Rightarrow x=-\frac{37}{24}\)

\(b)\) \(\frac{4}{9}:x=-\frac{1}{3}+1\frac{1}{6}\)

\(\Rightarrow\frac{4}{9}:x=-\frac{1}{3}+\frac{7}{6}\)

\(\Rightarrow\frac{4}{9}:x=-\frac{2}{6}+\frac{7}{6}\)

\(\Rightarrow\frac{4}{9}:x=\frac{5}{6}\)

\(\Rightarrow x=\frac{4}{9}:\frac{5}{6}\)

\(\Rightarrow x=\frac{4}{9}.\frac{6}{5}\)

\(\Rightarrow x=\frac{8}{15}\)

\(c)\left(3x-2\right)^3=-\frac{1}{27}\)

\(\Rightarrow3x-2=-\frac{1}{3}\)

\(\Rightarrow3x=-\frac{1}{3}+2\)

\(\Rightarrow3x=-\frac{1}{3}+\frac{6}{3}\)

\(\Rightarrow3x=\frac{5}{3}\)

\(\Rightarrow x=\frac{5}{3}:3\)

\(\Rightarrow x=\frac{5}{9}\)

d ) và e ) tự làm 

Chúc bạn học tốt !!! 

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\)

\(\Leftrightarrow-4x+13=6\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\frac{7}{4}\)

Vậy \(x=1\).

b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)

\(\Leftrightarrow-24x+37=10\)

\(\Leftrightarrow-24x=27\)

\(\Leftrightarrow x=\frac{9}{8}.\)

Mấy pài kia tương tự . :D

cậu khai triển các tích ra là ra thui mà cậu

a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)

Trả lời :.....................................

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125..........................\)

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Học sinh giỏi 6A

a, (x-2)^2 - (x-3)(x+3)=6

x^2-4x+4-(x^2-9)=6

x^2-4x+4-x^2+9=6

(x^2-x^2)-4x+13=6

-4x=-7

x=1,75

b, 4(x-3)^2 - (2x-1)(2x+1)=10

4(x^2-6x+9)-(4x^2-1)=10

4x^2-24x+36-4x^2+1=10

-24x+37=10

x=9/8

c,(x-4)^2 - (x+2)(x-2)=6

x^2-8x+16-(x^2-4)=6

x^2-8x+16-x^2+4=6

-8x+20=6

x=7/4

d, 9(x+1)^2 - (3x-2)(3x+2)=10

9(x^2+2x+1)-(9x^2-4)=10

9x^2+18x+9-9x^2+4=10

18x+13=10

x=-1/6

\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(-4x+13=6\)

\(-4x=6-13\)

\(-4x=-7\)

\(x=\frac{-7}{-4}\)

\(x=\frac{7}{4}\)

Vậy \(x=\frac{7}{4}\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)

\(4x^2-24x+36-4x^2+1=10\)

\(-24x+37=10\)

\(x=\frac{9}{8}\)

Vậy \(x=\frac{9}{8}\)

\(c,\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)

\(x^2-8x+16-\left(x^2-4\right)=6\)

\(x^2-8x+16-x^2+4=6\)

\(-8x+20=6\)

\(x=\frac{7}{4}\)

Vậy \(x=\frac{7}{4}\)

\(d,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)

\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)

\(9x^2+18x+9-9x^2+4=10\)

\(18x+13=10\)

\(x=\frac{-1}{6}\)

Vậy \(x=\frac{-1}{6}\)