a) Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)

\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b,\(< =>25x^2+10x+1-25x^2+9-30=0\)

\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)

c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)

\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)

\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)

\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)

a: Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

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

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

a) (3x + 2)² - (3x - 2)² = 5x + 38

<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2

b) 3(x - 2)² + 9(x - 1) = 3(x² + x - 3)

<=> 3x² - 12x + 12 + 9x - 9 = 3x² + 3x - 9

<=> -6x = -12 <=> x = 2

c) (x + 3)² - (x - 3)² = 6x + 8

<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3

d) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

<=> x³ - 3x² + 3x - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> 3x = -21 <=> x = -7

e) (x + 1)(x² - x + 1) - 2x = x(x - 1)(x + 1)

<=> x³ - 1 - 2x = x³ - x

<=> x = -1

a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)

\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)

\(\left(\sqrt{x-1}+\sqrt{x+3}+3\right)\left(\sqrt{x-1}+\sqrt{x+3}-2\right)=0\)

Đến đây em xét các trường hợp rồi bình phương lên là được nha

b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)

\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)

\(\left(\sqrt{3x-2}+\sqrt{x-1}-3\right)\left(\sqrt{3x-2}+\sqrt{x-1}+2\right)=0\)

Đến đây em xét các trường hợp rồi bình phương lên là được nha

a/ ĐKXĐ: $x\geq 1$

Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:

$a+b+2ab=6-(a^2+b^2)$

$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$

$\Leftrightarrow (a+b)^2+(a+b)-6=0$

$\Leftrightarrow (a+b-2)(a+b+3)=0$

Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$

$\Leftrightarrow a+b=2$

Mà $b^2-a^2=(x+3)-(x-1)=4$

$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$

$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$

$\Leftrightarrow x=1$ (tm)

b/

ĐKXĐ: $x\geq 1$

Đặt $\sqrt{3x-2}=a; \sqrt{x-1}=b(a,b\geq 0)$. Khi đó pt đã cho trở thành:

$a+b=a^2+b^2-6+2ab$

$\Leftrightarrow a^2+b^2+2ab-(a+b)-6=0$

$\Leftrightarrow (a+b)^2-(a+b)-6=0$

$\Leftrightarrow (a+b+2)(a+b-3)=0$

Hiển nhiên $a+b+2>0$ với mọi $a,b\geq 0$

Do đó $a+b-3=0\Leftrightarrow a+b=3$

$\Leftrightarrow b=3-a$.

Ta thấy $a^2-3b^2=1$. Thay $b=3-a$ vô thì:

$a^2-3(3-a)^2=1$

$\Leftrightarrow (a-2)(a-7)=0$

$\Leftrightarrow a=2$ hoặc $a=7$

Vì $a+b=3$ mà $a,b>0$ nên $a,b<3$. Do đó $a=2$

$\Leftrightarrow \sqrt{3x-2}=2$ 

$\Leftrightarrow x=2$ 

a: Ta có: \(\sqrt{\sqrt{x}+3}=4\)

\(\Leftrightarrow\sqrt{x}+3=16\)

\(\Leftrightarrow\sqrt{x}=13\)

hay x=169

b: Ta có: \(\sqrt{x+3}=\sqrt{1-5x}\)

\(\Leftrightarrow x+3=1-5x\)

\(\Leftrightarrow6x=-2\)

hay \(x=-\dfrac{1}{3}\left(nhận\right)\)

a) \(\sqrt{3+\sqrt{x}}=4\left(đk:x\ge0\right)\)

\(\Leftrightarrow3+\sqrt{x}=16\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\left(tm\right)\)

b) \(\sqrt{x+3}=\sqrt{1-5x}\left(đk:\dfrac{1}{5}\ge x\ge-3\right)\)

\(\Leftrightarrow x+3=1-5x\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\left(ktm\right)\)

Vậy \(S=\varnothing\)

c) \(\sqrt{x^2+6x+9}=3x-1\left(đk:x\ge\dfrac{1}{3}\right)\)

\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\)

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

\(\Leftrightarrow x+3=3x-1\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)

a. \(\sqrt{3+\sqrt{x}}=4\)      ĐKXĐ: \(x\ge0\)

<=> 3 + \(\sqrt{x}\) = 4²

<=> \(3+\sqrt{x}=16\)

<=> \(\sqrt{x}=16-3\)

<=> \(\sqrt{x}=13\)

<=> x = 13²

<=> x = 169 (TM)

b. \(\sqrt{x+3}=\sqrt{1-5x}\)           ĐKXĐ: \(x\ge\dfrac{1}{5}\)

<=> \(\left(\sqrt{x+3}\right)^2=\left(\sqrt{1-5x}\right)^2\)

<=> \(|x+3|=|1-5x|\)

<=> \(\left[{}\begin{matrix}x+3=1-5x\\-\left(x+3\right)=-\left(1-5x\right)\\x+3=-\left(1-5x\right)\\-\left(x+3\right)=1-5x\end{matrix}\right.\)

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

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

c. \(\sqrt{x^2+6x+9}=3x-1\)

<=> \(\sqrt{\left(x+3\right)^2}=3x-1\)

<=> \(|x+3|=3x-1\)

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

<=> \(\left[{}\begin{matrix}x+3=-3x=1\\-2x=-4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x=-2\\x=2\end{matrix}\right.\)

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

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)