Tim x biết \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
NN
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tim x\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x-1\right)=11\)
Tìm x biết
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
x=1 và 0
ms thỏa mản đề ra
=)))))))))))))))
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\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
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\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)
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Tim x:
1.\(\left(2x+1\right)\left(x-1\right)-x\left(2x-3\right)+3=0\)
2.\(\left(x^2+x-2\right)\left(x^2-x-2\right)-x^2\left(x^2-2\right)+8=0\)
\(1,\left(2x+1\right)\left(x-1\right)-x\left(2x-3\right)+3=0\)
\(\Rightarrow2x^2-2x+x-1-\left(2x^2-3x\right)+3=0\)
\(\Rightarrow2x^2-2x+x-1-2x^2+3x+3=0\)
\(\Rightarrow2x=-2\Rightarrow x=-1\)
\(2,\left(x^2+x-2\right)\left(x^2-x-2\right)-x^2\left(x^2-2\right)+8=0\)
\(\Rightarrow[\left(x^2\right)^2-\left(x-2\right)^2]-x^2\left(x^2-2\right)+8=0\)
\(\Rightarrow x^4-\left(x^2-4x+4\right)-x^4+2x^2+8=0\)
\(\Rightarrow x^4-x^2+4x-4-x^4+2x^2+8=0\)
\(\Rightarrow x^2+4x+4=0\)
\(\Rightarrow\left(x+2\right)^2=0\Rightarrow x=-2\)
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\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
Tim x
\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Rightarrow\left(2x+1\right)^4=\left(2x+1\right)^4.\left(2x+1\right)^2\)
\(\Rightarrow\left(2x+1\right)^2=\left(2x+1\right)^4:\left(2x+1\right)^4\)
\(\Rightarrow\left(2x+1\right)^2=1\)
\(\Rightarrow\left(2x+1\right)^2=1^2\)
\(\Rightarrow2x+1=1\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
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\(\left(2x+1\right)^4\)\(=\left(2x+1\right)^6\)
\(\Rightarrow\left(2x+1\right)^4\)\(=\left(2x+1\right)^4\)\(.\left(2x+1\right)^2\)
\(\Rightarrow\left(2x+1\right)^2\)\(=\left(2x+1\right)^4\)\(:\left(2x+1\right)^4\)
\(\Rightarrow\)\(\left(2x+1\right)^2\)\(=1^2\)
\(\Rightarrow\)\(2x+1=1\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
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\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Rightarrow\left(2x+1\right)^4=\left(2x+1\right)^4.\left(2x+1\right)^2\)
\(\Rightarrow\left(2x+1\right)^2=\left(2x+1\right)^4:\left(2x+1\right)^4\)
\(\Rightarrow\left(2x+1\right)^2=1\)
\(\Rightarrow\left(2x+1\right)^2=1^2\)
\(\Rightarrow2x+1=1\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
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Tìm x và y biết:d)-1frac{2}{3}-left(left|2xright|+frac{5}{6}right)-2e)left(-frac{1}{2}+frac{1}{3}right):left|1-2xright|-1frac{1}{4}:left(-frac{5}{8}right).left(-frac{1}{2}right)^2frac{1}{3}c)left|2x-1right|+left|2y+1right|+left|2x-yright|0b)left|2x-1right|2x-1a)left|x-3right|x+4
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Tìm x và y biết:
d)\(-1\frac{2}{3}-\left(\left|2x\right|+\frac{5}{6}\right)=\)\(-2\)e)\(\left(-\frac{1}{2}+\frac{1}{3}\right):\left|1-2x\right|-1\frac{1}{4}:\left(-\frac{5}{8}\right).\left(-\frac{1}{2}\right)^2=\frac{1}{3}\)
c)\(\left|2x-1\right|+\left|2y+1\right|+\left|2x-y\right|=0\)b)\(\left|2x-1\right|=2x-1\)
a)\(\left|x-3\right|=x+4\)
Bài 1 : Tìm GTNN của : \(A=\left|x+8\right|+\left|2x+7\right|+\left|3x+6\right|+\left|4x-7\right|+\left|3x-6\right|+\left|2x-7\right|+\left|x-8\right|-100\)
\(\left(x-1\right)^6=\left(x-1\right)^8\)
b) \(\left(2x+1\right)^4=\left(2x+1\right)^8\)
a/ (x - 1)6 = (x - 1)8
=> (x - 1)6 [1 - (x - 1)2] = 0
=> (x - 1)6 (1 - x2 + 2x - 1) = 0
=> (x - 1)6 (-x2 + 2x) = 0
=> x - 1 = 0 => x = 1
hoặc - x2 + 2x = 0 => x = 0 hoặc x = 2
Vậy x = 0, x = 1, x = 2
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Tim x
a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1-3x^2\right)=54\)
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1-3x2)=54
\(\Rightarrow\)x3+9x2+27x+27-x(9x2+6x+1)+(2x+1)(x2-2x+1)=54
\(\Rightarrow\)x3+9x2+27x+27-9x3-6x2-x+2x3-4x2+2x+x2-2x+1=54
\(\Rightarrow\)-6x3+26x+28=54
\(\Rightarrow\)-6x3+26x=54-28
\(\Rightarrow\)-6x3+26x=26
\(\Rightarrow\)-6x3+26x-26=0
\(\Rightarrow\)-2(3x3+13x+14)
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\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)\(\sqrt{ }\)\(\left|2x+3\right|+\left|2x-1\right|\)=\(\dfrac{8}{3.\left(x+1\right)^2+2}\)