1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)

\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)

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

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

=>x=-3(nhận) hoặc x=2(nhận)

2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

hay x=1(nhận)

c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)

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

\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)

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

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

=>x=-3(nhận) hoặc x=2(nhận)

a) ĐKXĐ: \(x\ne0\)

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

\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)

Suy ra: \(3x=10\)

\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)

b) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)

\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)

Suy ra: \(3x-4x=8\)

\(\Leftrightarrow-x=8\)

hay x=-8(thỏa ĐK)

Vậy: S={-8}

c)ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)

\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)

Suy ra: 2x+3x=10

\(\Leftrightarrow5x=10\)

hay x=2(thỏa ĐK)

Vậy: S={2}

d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)

\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)

\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)

\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)

\(\Leftrightarrow\) x = a (TM)

Vậy S = {a}

Chúc bn học tốt!

giúp mình bài ni với :3x^2(x+1)-5x(x+1)^2+4(x+1)

a)2x(8x-1)²(4x-1)=9

\(\Leftrightarrow\) (64x²-16x+1)(8x²-2x)=9

\(\Leftrightarrow\) 512x⁴-256x³+40x²-2x=9

\(\Leftrightarrow\) 512x⁴-256x³+40x²-2x-9=0

\(\Leftrightarrow\) 512x⁴-128x³-64x²-128x³+32x²+16x+72x²-18x-9=0

\(\Leftrightarrow\) (512x⁴-128x³-64x²)-(128x³-32x²-16x)+(72x²-18x-9)=0

\(\Leftrightarrow\) 64x²(8x²-2x-1)-16x(8x²-2x-1)+9(8x²-2x-1)=0

\(\Leftrightarrow\) (64x²-16x+9)(8x²-2x-1)=0

\(\Leftrightarrow\) (64x²-16x+9)(8x²-4x+2x-1)=0

\(\Leftrightarrow\) (64x²-16x+9)(2x-1)(4x+1)=0

\(\Rightarrow\left\{{}\begin{matrix}2x-1=0\\4x+1=0\end{matrix}\right.\) (Vì 64x²-16x+9>0)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\)\(\left[2x\left(4x+1\right)\right]\left(8x-1\right)^2=9\)

\(\Rightarrow\left(64x^2-16x+1\right)\left(8x^2-2x\right)=9\) (1)

đặt \(8x^2-2x=a\Rightarrow64x^2-16x=8a\)

từ đó (1)có dạng : (8a+1)a=9

\(\Rightarrow8a^2+a-9=0\)

\(\Rightarrow8a^2-8a+9a-9=0\)

\(\Rightarrow8a\left(a-1\right)+9\left(a-1\right)=0\)

\(\Rightarrow\left(8a+9\right)\left(a-1\right)=0\)

\(\left[\begin{matrix}8a+9=0\\a-1=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}a=\frac{-9}{8}\\a=1\end{matrix}\right.\)

từ đó thay vào tìm x

câu b bạn làm tương tự câu a mình làm trên

a: Ta có: \(3x+5\le4x-9\)

\(\Leftrightarrow-x\le-14\)

\(\Leftrightarrow x\ge14\)

b: Ta có: \(6-2x< 6-x\)

\(\Leftrightarrow-x< 0\)

hay x>0

c: Ta có: \(7\left(x-1\right)+5>-3x\)

\(\Leftrightarrow7x-7+5+3x>0\)

\(\Leftrightarrow10x>2\)

hay \(x>\dfrac{1}{5}\)

a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

=>\(x^3-x^2-4x^2+4x+4x-4=0\)

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

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

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

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

2: \(x^3+3x^2=x+6\)

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

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

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

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

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

\(2x+3\sqrt{x}=1\)

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

=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

=>\(x^4-3x^3+4x^2-3x+1=0\)

=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

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

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

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

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

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

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

b.

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

c.

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

ĐKXĐ: \(x\ge0\)

Do \(x\ge0\Rightarrow\left\{{}\begin{matrix}2x\ge0\\3\sqrt{x}\ge0\end{matrix}\right.\)

\(\Rightarrow2x+3\sqrt{x}+1>0\)

Pt đã cho vô nghiệm

d.

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

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

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\) chia cả 2 vế của pt cho \(x^2\)

\(\Rightarrow x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

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

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

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-2x+1=0\end{matrix}\right.\)

\(\Rightarrow x=1\)

a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)

Vậy: \(S=\left\{-1\right\}\)

b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)

Vậy: \(S=\left\{15\right\}\)

c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)

Vậy: \(S=\left\{5\right\}\)

d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)

Vậy: \(S=\left\{12\right\}\)

e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=12

=>x=4

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=36

=>x=12

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

a) \(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

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

Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)

Vì \(x^2+2\ge2>0\forall x\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)

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

Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)