Với giá trị nào của biến thì biểu thức sau có nghĩa:
a,\(\frac{2a}{a+1}\) b,\(\frac{x^2+1}{x-1}+\frac{1}{x}\)
c, \(\frac{2}{b^2+1}+3b\) d, \(\frac{5}{x\left(x^2\right)+1}\)
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a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
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!!!!
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)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
a) \(P=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x\left(x+2\right)+x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}:\frac{6}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{-6x}{6x\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{-1}{x+2}\)
b) Khi \(\left|x\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=-\frac{1}{\frac{3}{4}+2}=-\frac{4}{11}\\P=-\frac{1}{-\frac{3}{4}+2}=-\frac{4}{5}\end{cases}}\)
c) Để P = 7
\(\Leftrightarrow-\frac{1}{x+2}=7\)
\(\Leftrightarrow7\left(x+2\right)=-1\)
\(\Leftrightarrow7x+14=-1\)
\(\Leftrightarrow7x=-15\)
\(\Leftrightarrow x=-\frac{15}{7}\)
Vậy để \(P=7\Leftrightarrow x=-\frac{15}{7}\)
d) Để \(P\inℤ\)
\(\Leftrightarrow1⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-3;-1\right\}\)
a) Phân số\(\frac{2a}{a+1}\)có nghĩa
\(\Leftrightarrow a+1\ne0\)
\(\Leftrightarrow a\ne-1\)
Vậy Phân số\(\frac{2a}{a+1}\)có nghĩa \(\Leftrightarrow a\ne-1\)
c) Ta có: \(b^2\ge0\Rightarrow b^2+1\ge1\forall b\)
\(\Rightarrow\frac{2}{b^2+1}+3b\)luôn có nghĩa \(\forall x\in R\)