Tim so huu ti x, biet:
a/ \(5< 5^x< 625\)
b/ \(2^{x-1}=16\)
c/ \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)\(\left(x\in Z\right)\)
Những câu hỏi liên quan
cho 3 so huu ti phan biet c/m
\(\sqrt{\frac{1}{\left(x-4\right)^2}+\frac{1}{\left(y-z\right)^2}+\frac{1}{\left(z-x\right)^2}}\) la so huu ti
Chứng minh bằng biến đổi tương đương điều sau:
\(\left(\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{z-x}\right)^2=\frac{1}{\left(x-y\right)^2}+\frac{1}{\left(y-z\right)^2}+\frac{1}{\left(z-x\right)^2}\)
là có thể chứng minh được bài toán.
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tim so huu ti x biet :
a)\(\left(2x-3\right)^4=\left(2x-3\right)^6\)
b) \(\left(3x+5\right)^3=\left(3x+5\right)^{2016}\)
c) \(\left(2x+1\right)^{2015}=\left(2x+1\right)^{2017}\)
a, (2x-3)4=(2x-3)6
=> (2x-3)6 : (2x-3)4=1
=> (2x-3)3=
=> 2x-3=1
=> 2x=4
=> x=2
b, (3x+5)3=(3x+5)2016
=> (3x+5)2016 : (3x+5)3=1
=> (3x+5)2013=1
=> 3x+5=1
=> 3x=-4
=> x=-4/3
c, (2x+1)2015=(2x+1)2017
=> (2x+1)2017 : (2x+1)2015=1
=> (2x+1)2=1
=> 2x+1=1
=> 2x=0
=> x=0
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Tim x,y,z biet:
\(x+1=y+2=z+3và\left(x-\frac{1}{5}\right)\left(y+\frac{1}{3}\right)\left(z-6\right)=0\)
Bai 1:a)Tim x biet\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2009}{2011}\)
b)\(\left(x-1\right)\times f\left(x\right)=\left(x+4\right)\times f\left(x\right)\)voi moi x
Bai 2;Tim x;y;z biet a)\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\) b)\(\frac{2x+1}{5}=\frac{3y-z}{7}=\frac{2x+3y-1}{6x}\)
Rút gọn:a) dfrac{3left(x-yright)left(x-zright)^2}{6left(x-yright)left(x-zright)}b) dfrac{6x^2y^2}{8xy^5}c) dfrac{3xleft(1-xright)}{2left(x-1right)}d) dfrac{9-left(x+5right)^2}{x^2+4x+4}e) dfrac{x^2-2x+1}{x^2-1}f) dfrac{8x-4}{8x^3-1}g) dfrac{x^2+5x+6}{x^2+4x+4}k) dfrac{20x^2-45}{left(2x+3right)^2}
Đọc tiếp
Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
a: \(=\dfrac{x-z}{2}\)
b: \(=\dfrac{3x}{4y^3}\)
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\(choP=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)....a,tìm.x\in Z.để.P\in Z...b,tìm.x\in z.để.P\in z...c,tìm.x.để.\left|P\right|=P...d,tìm.x.để.\sqrt{P}>P\)
a) \(-5+|3x-1|+6=|-4|\)
b)\(\left(x-1\right)^2=\left(x-1\right)^4\)
c)\(5^{-1}.25^x=125\left(x\in Z\right)\)
d)\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
a) -5 + |3x - 1| + 6 = |-4|
=> -5 + |3x - 1| + 6 = 4
=> 1 + |3x - 1| = 4
=> |3x - 1| = 4 - 1
=> |3x - 1| = 3
=> \(\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}}\)
=> \(\orbr{\begin{cases}3x=4\\3x=-2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}\)
Vậy ...
d) |x + 1| + |x + 2| + |x + 3| = 4x
Ta có: |x + 1| \(\ge\)0 \(\forall\)x
|x + 2| \(\ge\)0 \(\forall\)x
|x + 3| \(\ge\)0 \(\forall\)x
=> |x + 1| + |x + 2| + |x + 3| \(\ge\)0 \(\forall\)x => 4x \(\ge\)0 \(\forall\) x=> x \(\ge\)0 \(\forall\)x
=> x + 1 + x + 2 + x + 3 = 4x
=> 3x + 6 = 4x
=> 6 = 4x - 3x
=> x = 6
Vậy...
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b) (x - 1)2 = (x - 1)4
=> (x - 1)2 - (x - 1)4 = 0
=> (x - 1)2 .[1 - (x - 1)2 ] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\)
=> \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy x = {1; 2; 0}
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Phan nguyen cua so huu ti x, ki hieu [x] la so nguyen lon nhat ko vuot qua x.
Hay tim phan nguyen cua: \(\left[\frac{5}{2}\right];\left[\frac{-3}{2}\right];\left[0,2\right]\)
tim x biet
\(\left(x-\frac{1}{3}\right).\left(y-\frac{1}{2}\right).\left(z-5\right)=0\)
và x+2=y+1=z+3
\(\left(x-\frac{1}{3}\right)\left(y-\frac{1}{2}\right)\left(z-5\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{2}\\z=5\end{cases}}\)
Vì \(z+3=y+1\Rightarrow y=7\)
Lại có \(y+1=x+2\Rightarrow x=8-2=6\)
Vậy x = 6 ; y = 7 ; z = 5
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