Tìm biểu thức Q biết: 5 x x 2 + 2 x + 1 . Q = x x 2 − 1
A. x + 1 x − 1
B. x − 1 x + 1
C. x − 1 5 ( x + 1 )
D. x + 1 5 ( x − 1 )
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Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
1. a, \(2^{x+2}.3^{x+1}.5^x=10800\)
\(2^x.2^2.3^x.3.5^x=10800\)
\(\Rightarrow\left(2.3.5\right)^x.12=10800\)
\(\Rightarrow30^x=\frac{10800}{12}=900\)
\(\Rightarrow30^x=30^2\)
\(\Rightarrow x=2\)
b,\(3^{x+2}-3^x=24\)
\(\Rightarrow3^x\left(3^2-1\right)=24\)
\(\Rightarrow3^x.8=24\)\(\Rightarrow3^x=3^1\Rightarrow x=1\)
2, c, Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Dấu bằng xảy ra khi \(ab\ge0\)
Ta có: \(\left|x-2017\right|=\left|2017-x\right|\)
\(\Rightarrow\left|x-1\right|+\left|2017-x\right|\ge\left|x-1+2017-x\right|\)\(=\left|2016\right|=2016\)
Dấu bằng xảy ra khi \(\left(x-1\right)\left(2017-x\right)\ge0\)\(\Rightarrow2017\ge x\ge1\)
Vậy \(Min_{BT}=2016\)khi \(2017\ge x\ge1\)
d, Áp dụng BĐT \(\left|a\right|-\left|b\right|\le\left|a-b\right|\forall a,b\inℝ\)
Dấu bằng xảy ra khi \(b\left(a-b\right)\ge0\)
Ta có \(B=\left|x-2018\right|-\left|x-2017\right|\le\left|x-2018-x+2017\right|\)
\(\Rightarrow B\le1\)
Dấu bằng xảy ra khi \(\left(x-2017\right)\left[\left(x-2018\right)-\left(x-2017\right)\right]\ge0\)
\(\Rightarrow x\le2017\)
Vậy \(Max_B=1\) khi \(x\le2017\)
để BT \(\frac{5}{\sqrt{2x+1}+2}\) nguyên thì \(\sqrt{2x+1}+2\inƯ\left(5\right)\)
suy ra \(\sqrt{2x+1}+2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\sqrt{2x+1}\in\left\{-7;-3;-1;3\right\}\)
Mà \(\sqrt{2x+1}\ge0\) nên \(\sqrt{2x+1}\)chỉ có thể bằng 3
\(\Rightarrow2x+1=9\Rightarrow x=4\)( thỏa mãn điều kiện \(x\ge-\frac{1}{2}\))
Đây là cách lớp 9. Mk đang phân vân ko biết giải theo cách lớp 7 thế nào!!!!
a) Ta có: A = \(\left(\frac{x}{x-1}+\frac{x}{x^2-1}\right):\left(\frac{2}{x^2}-\frac{2-x^2}{x^3+x^2}\right)\)
A = \(\left(\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2\left(x+1\right)}{x^2\left(x+1\right)}-\frac{2-x^2}{x^2\left(x+1\right)}\right)\)
A = \(\left(\frac{x^2+x+x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x+2-2+x^2}{x^2\left(x+1\right)}\right)\)
A = \(\left(\frac{x^2+2x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2+2x}{x^2\left(x+1\right)}\right)\)
A = \(\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x^2\left(x+1\right)}{x\left(x+2\right)}\)
A = \(\frac{x^2}{x+1}\)
b) ĐKXĐ: x \(\ne\)\(\pm\)1; x \(\ne\)0; x \(\ne\)-2
Ta có: A = 4
<=> \(\frac{x^2}{x+1}=4\)
<=> x2 = 4(x + 1)
<=> x2 - 4x - 4 = 0
<=>(x2 - 4x + 4) - 8 = 0
<=> (x - 2)2 = 8
<=> \(\orbr{\begin{cases}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\sqrt{2}+2\\x=2-2\sqrt{2}\end{cases}}\)(tm)
Vậy ...
c) Ta có: A < 0
<=> \(\frac{x^2}{x+1}< 0\)
Do x2 \(\ge\)0 => x + 1 < 0
=> x < -1
Vậy để A < 0 thì x < -1 và x khác -2
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
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