b) 9x^2-(x-4)^2+0

9X^2-(x-4)^2=0 nha

mình gõ nhầm

b1 A=10000 B=23386

b2 a,x=5 

b,x=4 x=2 

có lẽ bn nên tự lm thì hơn

a: \(x^2-10x=-25\)

\(\Leftrightarrow x-5=0\)

hay x=5

b: \(x^2-6x+8=0\)

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

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

a. \(x^4-16=0\\ \Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)=0\)

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

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

b. \(x^2-9x+8=0\\ \Leftrightarrow x^2-x-8x+8=0\\ \Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

a. x⁴ - 16 = 0
=> x⁴ = 16
=> x⁴ = 2⁴
=> x = 2
b. x² - 9x + 8 = 0
=> x² - 8x - x + 8 = 0
=> ( x² - x ) - ( 8x - 8 ) = 0
=> x(x-1) - 8(x-1)=0
=> (x-1)(x-8)=0
=>TH1: x-1=0       TH2 : x-8=0
=> x=1                       => x=8
 

a: \(x^4-2x^3-25x^2+50x=0\)

=>\(x^3\left(x-2\right)-25x\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x^3-25x\right)=0\)

=>x(x-2)(x^2-25)=0

=>x(x-2)(x+5)(x-5)=0

=>x∈{0;2;-5;5}

b: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>x∈{1;2}

c: \(9x^2-4-2\left(3x-2\right)^2=0\)

=>(3x-2)(3x+2)-(3x-2)(6x-4)=0

=>(3x-2)(3x+2-6x+4)=0

=>(3x-2)(-3x+6)=0

=>(x-2)(3x-2)=0

=>x∈{2;2/3}

d: \(9x^2+90x+225-\left(x-7\right)^2=0\)

=>\(\left(3x+15\right)^2-\left(x-7\right)^2=0\)

=>(3x+15-x+7)(3x+15+x-7)=0

=>(2x+22)(4x+8)=0

=>2(x+11)*4*(x+2)=0

=>(x+11)(x+2)=0

=>x∈{-11;-2}

e: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+3x+5\right)=0\)

mà \(x^2+3x+5=x^2+3x+\frac94+\frac{11}{4}=\left(x+\frac32\right)^2+\frac{11}{4}>0\forall x\)

nên x-2=0

=>x=2

g: (x+1)(x+2)(x+3)(x+4)-24=0

=>\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

=>\(\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24=0\)

=>\(\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

mà \(x^2+5x+10=x^2+5x+\frac{25}{4}+\frac{75}{4}=\left(x+\frac52\right)^2+\frac{75}{4}\ge\frac{75}{4}>0\forall x\)

nên \(x^2+5x=0\)

=>x(x+5)=0

=>x∈{0;-5}

a: \(x\left(1-2x\right)+2x^2=14\)

=>\(x-2x^2+2x^2=14\)

=>x=14

b: \(x\left(x-5\right)+3x-15=0\)

=>\(\left(x-5\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

<=> 2x(x + 2) = 1

Vì 1 chỉ có tích là 1 . 1 nên:

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

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

=>x-3=0

hay x=3

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

bạn ơi có câu c không bạn

1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow x=\dfrac{5}{2}\)

2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)

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

\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)

\(\Leftrightarrow x=\dfrac{1}{8}\)