tìm x
|x|-1=\(\frac{3}{2}\)
Những câu hỏi liên quan
Tìm x : \(\frac{2x-\frac{x-1}{2}}{3}-\frac{\frac{x+1}{2}-\frac{2x-3}{3}}{2}=\frac{\frac{x-1}{2}-1}{3}-\frac{x-3}{4}\)
Tìm x: \(x-\frac{\frac{x}{2}-\frac{x}{3}}{4}-\frac{x}{6}=\frac{2\left(1+x\right)}{3}-\frac{\frac{x}{3}+\frac{1-x^7}{4}}{2}\)
Tìm x: \(\frac{\frac{1}{2}-\frac{x+2}{3}}{2}-\frac{2}{3}\left(x+1\right)=\frac{1}{4}\left(1-2x\right)-\frac{\frac{1}{3}-\frac{1-x}{2}}{2}\)
<=> \(\frac{-2x-1}{12}\)-\(\frac{2x+2}{3}=\frac{1-2x}{4}-\frac{3x-1}{12}\)
<=>\(\frac{-2x-1-8x-8-3+6x+3x-1}{12}=0\)
<=> -x-13=0=> x=-13
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\(\frac{\frac{1}{2}-\frac{x+2}{3}}{2}-\frac{2}{3}\left(x+1\right)=\frac{1}{4}\left(1-2x\right)-\frac{\frac{1}{3}-\frac{1-x}{2}}{2}\)
<=>\(6.\left(\frac{1}{2}-\frac{x+2}{3}\right)-8.\left(x+1\right)=3\left(1-2x\right)-6.\left(\frac{1}{3}-\frac{1-x}{2}\right)\)
<=>3-2.(x+2)-8x-8=3-6x-2+3.(1-x)
<=>3-2x-4-8x-8=3-6x-2+3-3x
<=>-10x-9=-9x+4
<=>x=-13
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Tìm x: \(\frac{3\frac{1}{2}x-4}{6}+\frac{2+\frac{1}{4}x}{3}=1\frac{1}{4}x\left(x-1\right)-\frac{7-\frac{3}{4}x}{3}\)
cho biết : A= \(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right).\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x+2}\)
a, tìm đkxđ của A và rút gọn A
b, tính giá trị của A khi x=3
c, tìm giá trị nguyên của x để A có giá trị nguyên
\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
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1 Tìm x:
( \(3x-2\frac{1}{3}\)):( \(3\frac{1}{4}-5\frac{2}{3}+1\frac{4}{5}\)) = \(2-1\frac{1}{3}x\)
2. Tìm x:
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
3. Tìm x:
\(\left(1+3x\right)^2-3x\left(2x+6\right)=\left(4-3x\right)\left(x+3\right)-\left(2x-1\right)^2\)
1) \(x=\frac{99}{196}\)
2) \(x=-2\)
3) \(x\approx-0,59\)
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Tìm x: \(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=\frac{2x-\frac{10-7x}{3}}{2}-\left(x+1\right)\)
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=\frac{2x-\frac{10-7x}{3}}{2}-\left(x+1\right)\)
<=>\(2x-\frac{x}{2}+\frac{3+x}{4}=2x-\frac{10-7x}{3}-2\left(x+1\right)\)
<=>\(24x-6x+9+3x=24x-40+28x-24x-24\)
<=>\(21x+9=28x-64\)
<=>\(-7x=-73\)
<=>x=73/7
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Tìm x : \(\frac{x+1}{x^2-4}-\frac{5x}{2-x}=\frac{4}{x+2}+3\)
\(\Leftrightarrow x+1+5x\left(x+2\right)=4\left(x-2\right)+3x^2-12\)
\(\Leftrightarrow x+1+5x^2+10x-4x+8-3x^2+12=0\)
\(\Leftrightarrow2x^2+7x+21=0\)
\(\text{Δ}=7^2-4\cdot2\cdot21=49-168< 0\)
Vì Δ<0 nên phương trình vô nghiệm
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Tìm x : \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=2016-2016\)
\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)
\(\Rightarrow\left(x-2017\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\right)=0\)
Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)
=> x = 2017
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\(Q=\left(\frac{\sqrt{x}}{2+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
a, rút gọn
b, tìm Q < -1
c, tìm x để Q = \(\frac{-3}{4}\)
d, tìm x \(\in\)Z để Q \(\in\)Z