( x-1)2 = ( x -1 )4
( x - 5 ) 3 = ( x -5 )15
Đề bài là tìm x nha
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bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\)
\(x\times\dfrac{11}{16}=2\)
\(x=2:\dfrac{11}{16}\)
\(x=\dfrac{32}{11}\)
Bài 1 :
\(\dfrac{x}{16}\times\left(2017-1\right)=2\)
\(\dfrac{x}{16}\times2016=2\)
\(\dfrac{x}{16}=\dfrac{2}{2016}\)
\(x=\dfrac{2}{2016}\times16\)
\(x=\dfrac{1}{63}\)
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
Bài 1:
1 - 2 + 3 - 4 + .... + 2009 - 2010
= (1 - 2) + (3 - 4) + ... + (2009 - 2010)
= -1 . 1005
= -1005
Bài 2:
a) (15 - x) - (-x + 12) = 7 - (-5 + x)
=> 15 - x + x - 12 = 7 + 5 - x
=> -x + x + x = 7 + 5 - 15 + 12
=> x = 9
b) (x - 5)4 = (x - 5)6
=> x - 5 = 1 hoặc x - 5 = 0
=> x = 6 hoặc x = 5
c) (x + 1) + (x + 3) + ... + (x + 99) = 0
=> (x . 50) + (1 + 3 + ... + 99) = 0
=> (x . 50) + 2500 = 0
=> x . 50 = -2500
=> x = -50
\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\left(x-1\right)^2-\left(x-1\right)^2\left(x-1\right)^2=0\)
\(\left(x-1\right)^2\left[1-\left(x-1\right)^2\right]=0\)
\(\left(x-1\right)^2\left(1-x+1\right)\left(1+x-1\right)=0\)
\(\left(x-1\right)^2\left(2-x\right)x=0\)
\(\Rightarrow\left(x-1\right)^2=0\)hoac \(\Rightarrow\orbr{\begin{cases}x=0\\2-x=0\end{cases}}\)
\(\Rightarrow x-1=0\) hoac \(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(\Rightarrow x=1\)hoac \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(\left(x-5\right)^3=\left(x-5\right)^{15}\)
\(\left(x-5\right)^3-\left(x-5\right)^{15}=0\)
\(\left(x-5\right)^3-\left(x-5\right)^3\left(x-5\right)^{12}=0\)
\(\left(x-5\right)^3\left[1-\left(x-5\right)^{12}\right]=0\)
cảm ơn nha