a) \(\Leftrightarrow x^2=81\Leftrightarrow x=+-9\)

b) \(-x^2=-4\Leftrightarrow x=+-2\)

c) \(\frac{3x}{2,7}=\frac{1}{3}:2\frac{1}{4}\Leftrightarrow\frac{3x}{2,7}=\frac{4}{27}\Leftrightarrow3x=\frac{2}{5}\Leftrightarrow x=\frac{2}{15}\)

`@` ` \text {Ans}`

`\downarrow`

`a,`

`1/4+3/4*x=3/2-x`

`=> 1/4 + 3/4x - 3/2 + x = 0`

`=> (1/4 - 3/2) + (3/4x + x) = 0`

`=> -5/4 + 7/4x = 0`

`=> 7/4x = 5/4`

`=> x = 5/4 \div 7/4`

`=> x = 5/7`

Vậy, `x=5/7`

`b,`

`3/5*x-1/4=1/10*x-1/2`

`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`

`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`

`=> 1/2x + 1/4 = 0`

`=> 1/2x = -1/4`

`=> x = -1/4 \div 1/2`

`=> x = -1/2`

Vậy, `x=-1/2`

`c,`

`3x-3/5=x-1/4`

`=> 3x - 3/5 - x + 1/4 = 0`

`=> (3x - x) - (3/5 - 1/4) = 0`

`=> 2x - 7/20 = 0`

`=> 2x = 0,35`

`=> x = 0,35 \div 2`

`=> x = 7/40`

Vậy, `x=7/40`

`d,`

`3/2*x-2/5=1/3*x-1/4`

`=>  3/2x - 2/5 - 1/3x + 1/4 = 0`

`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`

`=> 7/6x - 3/20 = 0`

`=> 7/6x = 3/20`

`=> x = 3/20 \div 7/6`

`=> x = 9/70`

Vậy, `x=9/70`

`@` `\text {Kaizuu lv uuu}`

a. 9x² - 6x - 3 = 0

<=> 3(3x² - 2x - 1) = 0

<=> 3(3x² - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

<=> 3(3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

b. (2x + 1)² - 4(x + 2)² = 9

<=> (2x + 1)² - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

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

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

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

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

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

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: Ta có: \(7x+25=144\)

\(\Leftrightarrow7x=119\)

hay x=17

b: Ta có: \(33-12x=9\)

\(\Leftrightarrow12x=24\)

hay x=2

c: Ta có: \(128-3\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

hay x=31

d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)

\(\Leftrightarrow5\left(726-3x\right)=2175\)

\(\Leftrightarrow726-3x=435\)

\(\Leftrightarrow3x=291\)

hay x=97

e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)

\(\Leftrightarrow41-\left(2x+5\right)=18\)

\(\Leftrightarrow2x+5=23\)

\(\Leftrightarrow2x=18\)

hay x=9

Bn cần bài nào vậy

f: Ta có: \(10-4x+120:8=16+1\)

\(\Leftrightarrow4x=17-25=-8\)

hay x=-2

g: Ta có: \(x+9x+7x+5x=2244\)

\(\Leftrightarrow22x=2244\)

hay x=102

h: Ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

hay x=7

b. (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 4x + 16 - (x² - 1) = 16

<=> x² + 4x + 16 - x² + 1 - 16 = 0

<=> x² - x² + 4x = 16 - 16 - 1

<=> 4x = -1

<=> x = \(\dfrac{-1}{4}\)

\(a,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\\ \Leftrightarrow x=\dfrac{23}{24}\\ b,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

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

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

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

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