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Lời giải:
a)
\(f\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3x.\frac{1}{x}\left(x+\frac{1}{x}\right)\)
\(=\left(x+\frac{1}{x}\right)^3+3\left(x+\frac{1}{x}\right)\) với mọi $x\neq 0$
$\Rightarrow f(x)=x^3+3x$
b)
$2f(x)+f(\frac{1}{x})=\frac{4x^2+3}{x}(1)$
Cho $x\to \frac{1}{x}$ thì:
$2f(\frac{1}{x})+f(x)=\frac{4}{x}+3x(2)$
Lấy $2.(1)-(2)$ suy ra: $3f(x)=5x+\frac{2}{x}$
$\Rightarrow f(x)=\frac{5}{3}x+\frac{2}{3x}$ với mọi $x\neq 0$
a: TH1: x>=2
=>2x-4<=x+12
=>x<=16
=>2<=x<=16
TH2: x<2
=>4-2x<=x+12
=>-3x<=8
=>x>=-8/3
=>-8/3<=x<2
b: TH1: x>=1
BPT sẽ là \(\dfrac{x-1}{x+2}< 1\)
=>(x-1-x-2)/(x+2)<0
=>x+2<0
=>x<-2(loại)
TH2: x<1
BPT sẽ là \(\dfrac{1-x}{x+2}-1< 0\)
=>(1-x-x-2)/(x+2)<0
=>(-2x-1)/(x+2)<0
=>(2x+1)/(x+2)>0
=>x>-1/2 hoặc x<-2
=>-1/2<x<1 hoặc x<-2
Dự đoán điểm rơi: x=3 ; y =4;z =2
ÁP dụng AM-Gm ta có:
\(\dfrac{8}{xyz}+\dfrac{x}{9}+\dfrac{y}{12}+\dfrac{z}{6}\ge4\sqrt[4]{\dfrac{8}{9.12.6}}=\dfrac{4}{3}\)
\(\dfrac{2}{xy}+\dfrac{x}{18}+\dfrac{y}{24}\ge3\sqrt[3]{\dfrac{2}{18.24}}=\dfrac{1}{2}\)
\(\dfrac{2}{yz}+\dfrac{y}{16}+\dfrac{z}{8}\ge3\sqrt[3]{\dfrac{2}{16.8}}=\dfrac{3}{4}\)
\(\dfrac{2}{xz}+\dfrac{z}{6}+\dfrac{x}{9}\ge3\sqrt[3]{\dfrac{2}{6.9}}=1\)
\(\dfrac{13}{18}x+\dfrac{13}{24}y\ge2\sqrt{\dfrac{169}{18.24}xy}\ge\dfrac{13}{3}\)
\(\dfrac{13}{24}z+\dfrac{13}{48}y\ge2\sqrt{\dfrac{169}{24.48}.yz}\ge\dfrac{13}{6}\)
Cộng tất cả theo vế ,ta thu được Đpcm.
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)
\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)
Bình phương 2 vế rút gọn
\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)
Đặt \(\sqrt{x^4-x^2}=a\)
\(\Rightarrow a^2-4\sqrt{3}a+12=0\)
\(\Leftrightarrow a=2\sqrt{3}\)
\(\Leftrightarrow x^4-x^2=12\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)
\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)
\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)
\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)
Toán lớp 8 nha