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a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)
b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)
c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)
d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)
a)
\(A=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\\ =\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\\ =\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Ta có: \(A=\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}}-\frac{1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
Với mọi a>0 và a≠1 ta có \(\sqrt{a}>0\Leftrightarrow\frac{1}{\sqrt{a}}>0\)
\(\Rightarrow A=1-\frac{1}{\sqrt{a}}< 1\left(đpcm\right)\)
c)
\(A=1-\frac{1}{\sqrt{a}}=\frac{1}{2}\Leftrightarrow\frac{1}{\sqrt{a}}=\frac{1}{2}\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\left(tm\right)\)
Vậy.......
a) A= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{2}\right)\) (x ≥ 0; x ≠ 4)
= \(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{2}\)
=\(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)
=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)
= \(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)
=\(\frac{2}{x+\sqrt{x}+1}\)
b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0
⇒x+\(\sqrt{x}\)+1 ≥ 1 > 0
mà 2 > 0
⇒ A > 0 (1)
Ta có:
\(x+\sqrt{x}+1\) ≥ 1
⇒ \(\frac{1}{x+\sqrt{x}+1}\) ≤ 1
⇒\(\frac{2}{x+\sqrt{x}+1}\) ≤ 2
⇒A ≤ 2 (2)
Từ (1) và (2) => 0 < A ≤ 2
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
b: \(=\left|b\cdot\left(b-1\right)\right|=b\cdot\left|b-1\right|\)
c: \(=\left|a\right|\cdot\left|a+1\right|=a\left(a+1\right)=a^2+a\)
d: \(=1-2a-4a=-6a+1\)
Lời giải:
a)
\(\sqrt{36(b-2)^2}=\sqrt{6^2(b-2)^2}=6\sqrt{(b-2)^2}=6|b-2|=6(2-b)\) do \(b<2\)
b)
\(\sqrt{b^2(b-1)^2}=\sqrt{b^2}\sqrt{(b-1)^2}=|b||b-1|\)
Do \(b< 0\Rightarrow b,b-1< 0\)
\(\Rightarrow \sqrt{b^2(b-1)^2}=|b||b-1|=-b(1-b)=b(b-1)\)
c) \(\sqrt{a^2(a+1)^2}=\sqrt{a^2}\sqrt{(a+1)^2}=|a||a+1|\)
\(=a(a+1)\) do \(a>0\)
d) \(\sqrt{(2a-1)^2}-4a=|2a-1|-4a\)
Vì \(a< \frac{1}{2}\Rightarrow 2a-1< 0\)
\(\Rightarrow \sqrt{(2a-1)^2}-4a=|2a-1|-4a=(1-2a)-4a=1-6a\)
\(M=\left(\frac{3}{\sqrt{a+1}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}+1\right)\)
\(=\left(\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\right):\left(\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1+a\right)\left(1-a\right)}}\right)\)
\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}.\frac{\sqrt{\left(1+a\right)\left(1-a\right)}}{3+\sqrt{\left(1-a\right)\left(1+a\right)}}\)
\(=\sqrt{1-a}\left(đpcm\right)\)
Bài 1:
ĐK: $a\geq 0; a\neq 1$
a)
\(P=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)
\(=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)\)
\(=(\sqrt{a}+1)^2(\sqrt{a}-1)^2=(a-1)^2\)
b) \(P< 7-4\sqrt{3}\)
\(\Leftrightarrow (a-1)^2< (2-\sqrt{3})^2\)
\(\Leftrightarrow \sqrt{3}-2< a-1< 2-\sqrt{3}\)
\(\Leftrightarrow \sqrt{3}-1< a< 3-\sqrt{3}\)
Vậy $\sqrt{3}-1< a< 3-\sqrt{3}$ và $a\neq 1$
Bài 2:
a)
\(A=\frac{2}{a-\sqrt{a}}.\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}=\frac{2(\sqrt{a}-1)^2}{\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}=\frac{2(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}+1)}\)
b)
Xét hiệu \(A-1=\frac{2\sqrt{a}-2-a-\sqrt{a}}{\sqrt{a}(\sqrt{a}+1)}=-\frac{a-\sqrt{a}+2}{\sqrt{a}(\sqrt{a}+1)}\)
Thấy rằng: \(a-\sqrt{a}+2=(\sqrt{a}-\frac{1}{2})^2+\frac{7}{4}>0; \sqrt{a}(\sqrt{a}+1)>0 \) với mọi $a>0; a\neq 1$ nên:
\(A-1=-\frac{a-\sqrt{a}+2}{\sqrt{a}(\sqrt{a}+1)}<0\Rightarrow A< 1\)
Với 0 < a < 1 ta có:
P = 1 + a 1 + a − 1 − a + 1 − a 2 1 − a 1 + a − 1 − a 2 1 − a 2 a 2 − 1 a = 1 + a 1 + a − 1 − a + 1 − a 2 1 − a 1 + a − 1 − a ( 1 − a ) ( 1 + a ) a 2 − 1 a = 1 + a 1 + a − 1 − a + 1 − a 1 + a − 1 − a 1 − a . 1 + a a 2 − 1 a = 1 + a + 1 − a 1 + a − 1 − a . 2 1 − a . 1 + a − ( 1 − a ) − ( 1 + a ) 2 a = 1 + a + 1 − a 1 + a − 1 − a . − 1 + a − 1 − a 2 2 a = − 1 + a + 1 − a 1 + a − 1 − a 2 a = − 1 + a − 1 + a 2 a = − 2 a 2 a = − 1