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a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)
=>-x^2+2x-1=10x-5x^2-11x-22
=>-x^2+2x-1=-5x^2-x-22
=>4x^2+3x+21=0
=>PTVN
b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)
=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)
=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80
=>20x+16=32x-80
=>-12x=-96
=>x=8
c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)
=>6x-18+7x-35=13x+4
=>-53=4(loại)
d: =>3(2x-1)-5(x-2)=3(x+7)
=>6x-3-5x+10=3x+21
=>3x+21=x+7
=>x=-7
e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1
=>-9x^2+9x-9=-9x^2+1
=>9x=10
=>x=10/9
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
1: =>3x+1=4
=>3x=3
hay x=1
2: \(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^9}{98^3}=\dfrac{1}{2^3}+\dfrac{7^9}{7^6\cdot2^3}\)
\(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^3}{2^3}=\dfrac{344}{2^3}\)
\(\Leftrightarrow x^2=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
3: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\x-\dfrac{2}{9}=-\dfrac{4}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{9}\end{matrix}\right.\)
4: =>x+2=0 và y-1/10=0
=>x=-2 và y=1/10
a: \(\Leftrightarrow\dfrac{5}{2}:\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{15}{4}-3=\dfrac{3}{4}\)
\(\Leftrightarrow\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{5}{2}:\dfrac{3}{4}=\dfrac{5}{2}\cdot\dfrac{4}{3}=\dfrac{20}{6}=\dfrac{10}{3}\)
=>3/4x+1/2=10/3 hoặc 3/4x+1/2=-10/3
=>3/4x=17/6 hoặc 3/4x=-23/6
=>x=34/9 hoặc x=-46/9
b: \(\Leftrightarrow\dfrac{9}{4}:\left|x+\dfrac{1}{3}\right|=6.5-2=\dfrac{9}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{9}{4}:\dfrac{9}{2}=\dfrac{1}{2}\)
=>x+1/3=1/2 hoặc x+1/3=-1/2
=>x=1/6 hoặc x=-5/6
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
bai 1
a) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|-3,75=-2,,15\)
\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=1,6\\x+\dfrac{4}{15}=-1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)
Vậy ....
b) \(\left|\dfrac{5}{3}x\right|=\left|-\dfrac{1}{6}\right|\)
\(\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\-1\end{matrix}\right.\)
bai 2
a) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{1}{4}-\left|y\right|\)
\(\left|\dfrac{1}{6}+x\right|=\dfrac{1}{4}-\left|y\right|\) (*)
với mọi x ta luôn có \(\left|\dfrac{1}{6}+x\right|\ge0\)
\(\Rightarrow\dfrac{1}{4}-\left|y\right|\ge0\)
\(\Rightarrow\left|y\right|\le\dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{4}-\left|y\right|=\left|\dfrac{1}{4}-y\right|\)
Nên từ * \(\Rightarrow\left|\dfrac{1}{6}+x\right|=\left|\dfrac{1}{4}-y\right|\)
\(\Rightarrow\left|\dfrac{1}{6}+x\right|-\left|\dfrac{1}{4}-y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{6}+x=0\\\dfrac{1}{4}-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left|x-y\right|+\left|y+25\right|=0\)
với mọi x, y tao luôn có \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+25\right|\ge0\end{matrix}\right.\)
mà \(\left|x-y\right|+\left|y+25\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+25\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=-25\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-25\\y=-25\end{matrix}\right.\)