1,Cho biểu thức A=( x-1 )2 -3
a,Tính giá trị biểu thức A biết : x = -2; 4; -5; 0.
b, Tìm giá trị nhỏ nhất của A.
K
Khách
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Những câu hỏi liên quan
10 tháng 3 2021
Bài 1 :
\(N=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)
hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)
mà \(xyz=2\Rightarrow-xyz=-2\)
hay N nhận giá trị -2
Bài 2 :
\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)
hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)
hay biểu thức trên nhận giá trị là 24
c, Ta có : \(a-b=3\Rightarrow a=3+b\)
hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)
\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi
NC
10 tháng 3 2021
1.Ta có:\(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)
\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)
2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)
Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)
Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)
Vậy....
31 tháng 10 2021
\(a,x=2\Leftrightarrow A=3\cdot4-4\cdot2-1=12-8-1=3\\ b,B=x^3-1-2x+x^2-2+x-x^3=x^2-x-3\\ c,C=B-A=x^2-x-3-3x^2+3x+1=-2x^2-2x-2\\ C=-2\left(x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\right)=-2\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{2}\le-\dfrac{3}{2}\\ C_{max}=-\dfrac{3}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
20 tháng 6 2021
a)
A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)
\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
MTC: 5(x-1)(x+1)
\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)
\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)
\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)
\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)
\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)
\(\Leftrightarrow10x+10\)
DT
1
7 tháng 2 2021
a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)
\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\cdot\dfrac{-\left(x-1\right)\left(x+1\right)}{2}\)
\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{2\left(x+1\right)}\)
\(=\dfrac{2\left(x-2\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2\cdot\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-\left(x^2-2x+1\right)\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-\left(x^3+2x^2-2x^2-4x+x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-\left(x^3-3x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-x^3+3x-2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)
