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a)\(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\left(ĐKXĐ:x\ne\pm\dfrac{2}{3}\right)\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}-\dfrac{6}{3x+2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+2\right)^2-6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow9x^2-6x+16-9x^2=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\dfrac{8}{3}\) (thỏa mãn ĐKXĐ)
Vậy .................
b) \(\dfrac{5-x}{4x^2-8x}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\left(ĐKXĐ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{2\left(5-x\right)+7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{4\left(x-1\right)+x}{8x\left(x-2\right)}\)
\(\Rightarrow10-2x+7x-14=4x-4+x\)
\(\Leftrightarrow5x-4=5x-4\)
\(\Leftrightarrow0x=0\) (vô số nghiệm)
Vậy \(S=R\backslash\left\{0;2\right\}\)
bt2.
A=[2(4x^2+4x+5)-2]/(4x^2+4x+5)
=2-2/[(4x+1)^2+4]
A>=2-2/4=3/2
khi x=-1/4
a: \(\Leftrightarrow x^2-4-x+4=-2\left(x^2-6x+8\right)\)
=>-2x^2+12x-16=x^2-x
=>-3x^2+13x-16=0
=>3x^2-13x+16=0
Δ=(-13)^2-4*3*16=169-192<0
=>PTVN
b: \(\Leftrightarrow\left(x-2\right)\left(x-3\right)+\left(x+1\right)\left(x+3\right)=2x^2+6\)
=>x^2-5x+6+x^2+4x+3=2x^2+6
=>-x+9=6
=>-x=-3
=>x=3(loại)
c: \(\Leftrightarrow4\left(2x-1\right)-3\left(x-6\right)=6\left(3x-2\right)\)
=>8x-4-3x+12=18x-12
=>-5x+8=18x-12
=>-23x=-20
=>x=20/23
d: \(\Leftrightarrow5\left(3x-1\right)+8x-21=12\left(x+2\right)-40\)
=>15x-5+8x-21=12x+24-40
=>23x-26=12x-16
=>11x=10
=>x=10/11
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
Nhớ tick cho mình nha
\(\dfrac{1}{3}\)x2 + \(\dfrac{1}{x^2}\) - 8x + 32 = \(\dfrac{1}{x^2}\) - 2x + 8 ĐK: x ≠ 0
⇔\(\dfrac{1}{3}\)x2 + \(\dfrac{1}{x^2}\) - \(\dfrac{1}{x^2}\) - 8x + 2x + 32 - 8 = 0
⇔\(\dfrac{1}{3}\)x2 - 6x +24 = 0
⇔\(\left(x-12\right)\) \(\left(x-6\right)\) = 0
⇔\(\left[{}\begin{matrix}x-12=0\\x-6=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=12\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
⇒ S = \(\left\{12;6\right\}\)