Mastered Ultra Instinct đề là tìm x thế ko có vế phải à???

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. 

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x=2x^3-16\)

<=>\(8x=-16\)

<=>\(x=-2\)

i. 

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>\(2x^2-x-3=2x^2+9x-5\)

<=>10x=2

<=>\(x=\dfrac{1}{5}\)

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

<=>

<=>

<=>\(8x=-16\)

<=>x=-2

i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(9x+6=0\)

<=>x=\(\dfrac{-2}{3}\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>

<=>10x=2

<=>

d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)

\(=8x^2+12x-8x^2-32x\)

=-20x

e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)

\(=10x^2+4x+6x^2-2x-9x+3\)

\(=16x^2-7x+3\)

f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)

\(=4x^2+x-4\)

len google di ban

mk chua hoc bai nay

a

a

a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)

=3(x-y)^2+9

b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)

c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)

a: (2x-3/2)(|x|-5)=0

=>2x-3/2=0 hoặc |x|-5=0

=>x=3/4 hoặc |x|=5

=>\(x\in\left\{\dfrac{3}{4};5;-5\right\}\)

b: x-8x^4=0

=>x(1-8x^3)=0

=>x=0 hoặc 1-8x^3=0

=>x=1/2 hoặc x=0

c: x^2-(4x+x^2)-5=0

=>x^2-4x-x^2-5=0

=>-4x-5=0

=>x=-5/4

a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha

b: \(=2x^2-3x+10x-15=2x^2+7x-15\)

a) \(8x^2+27=\left(x-1\right)^3+\left(x+4\right)^3\)

\(\Leftrightarrow8x^3+27=x^3-2x^2+x-x^2+2x-1+x^3+8x^2+16x+4x^2+32x+64\)

\(\Leftrightarrow8x^3+27=2x^3+9x^2+51x+63\)

\(\Leftrightarrow8x^3+27-2x^3-9x^2-51x-63=0\)

\(\Leftrightarrow6x^3-36-9x^2-51x=0\)

\(\Leftrightarrow3\left(2x^3-12-3x^2-17x\right)=0\)

\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)

\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)

\(\Leftrightarrow3\left[x\left(2x+3\right)-4\left(2x+3\right)\right]\left(x+1\right)=0\)

\(\Leftrightarrow3\left(2x+3\right)\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+3=0\\x-4=0\\x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)

tớ tưởng áp dụng công thức: \(\left(A+B\right)^3=A^3+B^3+3AB\left(A+B\right)\)

và \(\left(A-B\right)^3=A^3-B^3-3AB\left(A-B\right)\)

à hiểu rồi