\(\left(\text{|}x\text{|}-1,5\right)\left(x+1\right)=0\)
Các bạn giúp mình bài toán sau
\(\left(x+2\right)^3\text{-}\left(x+1\right)\left(x^2\text{-}x+1\right)=10\)
\(\left(x\text{-}1\right)^3\text{-}\left(x\text{-}2\right)\left(x^2+x+4\right).3x\left(x+1\right)=0\)
\(\left(x\text{-}3\right)^2\text{-}\left(x\text{-}2\right)\left(x^2+2x+4\right)\text{-}9x\left(x\text{-}1\right)=0\)
đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)
\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)
TH1: \(x-1=0\Leftrightarrow x=1\)
TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
tìm x, biết:
a) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
b) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
Tìm x, biết:
a) \(x\left(x-1\right)-x^2+2\text{x}=5\)
b) \(8\left(x-2\right)-2\left(3\text{x}-4\right)=2\)
c) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
d) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
Giải các phương trình sau:
a \(\left(x+2\right)\left(x+\text{4}\right)\left(x+6\right)\left(x+8\right)+16=0\)
b \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
c \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=0\)
d \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
\(Cho\text{ }x,y,z\text{ }\in R\text{ thỏa}\text{ }xyz=1.\text{Tìm Min:}\)
\(P=\left(\left|xy\right|+\left|yz\right|+\left|zx\right|\right)\left[15\sqrt{x^2+y^2+z^2}-7\left(x+y-z\right)\right]+1\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
\(\left(x+1\right)^4-2\left(x+3\right)\left(x+1\right)^2+12\text{x}=0\)0
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+2x-5\right)=0\)
bạn tách từng câu ra mik suy nghĩ từng câu
Tập nghiệm của bất phương trình -x2 + 4x - 3 < 0 là:
A. \(\left(-\infty;1\right)\cup\left(3;+\infty\right)\) B. (1; 3) C. \(\text{∀}\text{x}\in\text{R}\) D. \(\left(-1;1\right)\)