Tìm GTNN của các biểu thức:
a, \(P=2x+\frac{1}{\left(x-1\right)^2}\) với x > 1
b, \(P=2x+\frac{x^2}{x+1}\)với x > 0
c, \(P=\frac{4x}{1-x}+\frac{1}{x}\)với 0 < x < 1
d, \(P=\frac{\left(a+b\right)^2}{a-b}\)với a > b và ab = 1
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b, ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x+3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x\ne-3\end{matrix}\right.\)
- Ta có : \(\frac{x-3}{x+1}-\frac{x+1}{x+3}=\frac{x^2-x-10}{x^2+4x+3}\)
=> \(\frac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x+1\right)}{\left(x+3\right)\left(x+1\right)}=\frac{x^2-x-10}{\left(x+1\right)\left(x+3\right)}\)
=> \(\left(x-3\right)\left(x+3\right)-\left(x+1\right)\left(x+1\right)=x^2-x-10\)
=> \(x^2-9-x^2-2x-1-x^2+x+10=0\)
=> \(-x-x^2=0\)
=> \(x\left(x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
=> x = 0 .
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne3\\x\ne-1\end{matrix}\right.\)
Ta có : \(\frac{x}{2x-6}+\frac{x}{2x+2}+\frac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{4x}{\left(x+1\right)\left(x-3\right)}=0\)
=> \(x\left(x+1\right)+x\left(x-3\right)-4x=0\)
=> \(2x^2-6x=0\)
=> \(x\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> x > -1
1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
Đây là bài tìm GTNN mà đâu phải BĐT (BĐT mình hơi ngu).