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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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a) \(2^x=8\)
⇔ \(2^x=2^3\)
⇒ \(x=3\)
b) \(3^x=27\)
⇔ \(3^x=3^3\)
⇒ \(x=3\)
c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)
d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)
d) \(\left(x+1\right)^3=-125\)
⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)
⇔ \(x+1=-5\)
⇔ \(x=-5-1=-6\)
2:
a: (x-1,2)^2=4
=>x-1,2=2 hoặc x-1,2=-2
=>x=3,2(loại) hoặc x=-0,8(loại)
b: (x-1,5)^2=9
=>x-1,5=3 hoặc x-1,5=-3
=>x=-1,5(loại) hoặc x=4,5(loại)
c: (x-2)^3=64
=>(x-2)^3=4^3
=>x-2=4
=>x=6(nhận)
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1) \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)
\(\Leftrightarrow\) \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1
\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0
Ta thấy: 1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0
\(\Leftrightarrow\)x + 2009 = 0
\(\Leftrightarrow\)x = -2009
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Vì `x^4>=0,x^2>=0`
`=>x^4+x^2+1>=1>0`
`=>|x^4+x^2+1|=x^4+x^2+1`
`=>x^4+x^2+1=x^2-x+1`
`=>x^2=-x`
`=>x(x+1)=0`
`=>[(x=0),(x+1=0):}`
`=>[(x=0),(x=-1):}`
Vậy `x=0` hoặc `x=-1`
Ta có: \(\left|x^4+x^2+1\right|=x^2-x+1\)
\(\Leftrightarrow x^4+x^2+1-x^2+x-1=0\)
\(\Leftrightarrow x^4+x=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
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Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
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1) Ta có\(\frac{x+2}{5}=\frac{1}{x-2}\)
=> (x + 2)(x - 2) = 5
=> x2 + 2x - 2x - 4 = 5
=> x2 - 4 = 5
=> x2 = 9
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
2) \(\frac{3}{x-4}=\frac{x+4}{3}\)
=> (x - 4)(x + 4) = 9
=> x2 + 4x - 4x - 16 = 9
=> x2 - 16 = 9
=> x2 = 25
=> \(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
a, \(\frac{x+2}{5}=\frac{1}{x-2}ĐK:x\ne2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{5\left(x-2\right)}=\frac{5}{5\left(x-2\right)}\Leftrightarrow\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^2-2x+2x-4=5\Leftrightarrow x^2=9\Leftrightarrow x\pm3\)
b, \(\frac{3}{x-4}=\frac{x+4}{3}ĐK:x\ne4\)
\(\Leftrightarrow\frac{9}{\left(x-4\right)3}=\frac{\left(x+4\right)\left(x-4\right)}{3\left(x-4\right)}\Leftrightarrow9=x^2-4x+4x-16\)
\(\Leftrightarrow x^2-16=9\Leftrightarrow x^2=25\Leftrightarrow x=\pm5\)
c, \(\frac{x+2}{x+6}=\frac{3}{x}=1ĐK:x\ne0;-6\)
Xét : \(\frac{x+2}{x+6}=1\Leftrightarrow x+2=x+6\Leftrightarrow-4\ne0\)
Xét : \(\frac{3}{x}=1\Leftrightarrow3=x\)
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a)\(\Leftrightarrow\left[{}\begin{matrix}x^3-x-1=x^3+x+1\\x^3-x-1=-x^3-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(x+1\right)=0\\2x^3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
câu b) tương tự
b) Ta có: \(\left|x^4+x^2+1\right|=x^2+x+1\)
\(\Leftrightarrow x^4+x^2+1=x^2+x+1\)
\(\Leftrightarrow x^4-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
|x^4+x^2+1| = |x^2+x+1|
=> x^4+x^2+1 = x^2+x+1
=> x^4+x^2+1-x^2-x-1 = 0
=> x^4-x = 0
=> x.(x^3-1) = 0
=> x=0 hoặc x^3-1=0
=> x=0 hoặc x=1
Vậy .........