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\(\sin39^013'=0,6322\)
\(\cos52^018'=0,6115\)
\(\tan13^020'=0,2370\)
\(\cot10^017'=5,5118\)
\(\sin54^0=0,8090\)
\(\cos45^0=0,7071\)
a) để A xát định thì
\(\left[{}\begin{matrix}2x+10\ne0\\x\ne0\\2x\left(x-5\right)\ne0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}2x\ne-10\\x\ne0\\\left[{}\begin{matrix}2x\ne0\\x-5\ne0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x\ne-5\\x\ne0\\\left[{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x\ne0\\x\ne-5\\x\ne5\end{matrix}\right.\) thì A được xác định
a/ ĐKXĐ: x khác -1
\(P=\left(\dfrac{4}{x+1}-1\right):\dfrac{9-x^2}{x^2+2x+1}=\left(\dfrac{4}{x+1}-\dfrac{x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(3-x\right)\left(3+x\right)}\)
\(=\dfrac{3-x}{x+1}\cdot\dfrac{\left(x+1\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{x+1}{x+3}\)
b/ |x + 1| = 2
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
Với x = 1 P = \(\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
c/ \(\dfrac{x+1}{x+3}=\dfrac{x+3-2}{x+3}=\dfrac{x+3}{x+3}-\dfrac{2}{x+3}=1-\dfrac{2}{x+3}\)
ĐỂ P nguyên thì \(\dfrac{2}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(2\right)\)
\(x+3=\left\{-2;-1;1;2\right\}\)
=> \(x=\left\{-5;-4;-2;-1\right\}\) (tm)
Vậy............
ĐKXĐ: x>=0; x<>1
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:
\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)
\(=\dfrac{2-\sqrt{3}+\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{2}{2\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{2}\)
\(a,\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\left(\sqrt{x}-1\right)\)
\(=\dfrac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\)
\(=\sqrt{x}+1\)
\(b,\) Thay \(x=4-2\sqrt{3}\) vào biểu thức trên, ta được:
\(\sqrt{4-2\sqrt{3}}+1\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}+1\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+1\)
\(=\left|\sqrt{3}-1\right|+1\)
\(=\sqrt{3}-1+1\)
\(=\sqrt{3}\)
Vậy: ...
\(\text{#}Toru\)
\(a\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\\ =\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{x+\sqrt{x}+2}{x-1}\right).\sqrt{x}-1\\ =\dfrac{x+\sqrt{2}+1}{x-1}.\sqrt{x}-1\\ =\sqrt{x}+1\\ b,tacóx=4-2\sqrt{3}=\left(\sqrt{3}-\sqrt{1}\right)^2thãy=\sqrt{3}-\sqrt{1}vàobiểuthức,tađược\\ \sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-1=\sqrt{3}-1-1=\sqrt{3}-2\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
a, \(A=\dfrac{x-5}{x-3}-\dfrac{2x}{x+3}-\dfrac{2x^2-x+15}{9-x^2}\)
\(=\dfrac{\left(x-5\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-2x-15-2x^2+6x+2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x+3}\)
b, \(\left|x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Thay x =- 1 vào biểu thức A ,có :
\(\dfrac{-1}{-1+3}=\dfrac{-1}{2}\)
Vậy tại x = -1 gtri của bt A là -1/2
Vậy tại x = 3 biểu thức A ko có giá trị
c,\(\dfrac{x}{x+3}=\dfrac{x+3-3}{x+3}=1-\dfrac{3}{x+3}\)
Để A có giá trị nguyên
\(\Leftrightarrow\dfrac{3}{x+3}\) là số nguyên
\(\Leftrightarrow3⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
| \(x+3\) | 1 | -1 | 3 | -3 |
| x | -2 (t/m) | -4(t/m) | 0 (t/m) | -6(t/m) |
Vậy \(x\in\left\{0;-2;-4;-6\right\}\) thì A có giá trị nguyên
Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
a: \(A\simeq0.1787\)
b: \(B\simeq0.2582\)