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\(\Rightarrow B=\frac{\sqrt{b}\left(\sqrt{ab}-b\right)-\sqrt{a}\left(a-\sqrt{ab}\right)}{\left(a-\sqrt{ab}\right)\left(\sqrt{ab}-b\right)}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{b\sqrt{a}-\sqrt{b}^3-\sqrt{a}^3+a\sqrt{b}}{a\sqrt{ab}-ab-ab+b\sqrt{ab}}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{\left(b\sqrt{a}+a\sqrt{b}\right)-\left(\sqrt{a}^3+\sqrt{b}^3\right)}{a\sqrt{ab}-2ab+b\sqrt{ab}}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{\sqrt{ab}\left(\sqrt{b}+\sqrt{a}\right)-\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{ab}\left(a-2\sqrt{ab}+b\right)}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{ab}-a+\sqrt{ab}-b\right)}{\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)^2}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=-\frac{\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)^2}-\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{-\sqrt{a}-\sqrt{b}-ab\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
Tớ làm tới đây thui
rut gon
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left[1+\frac{\sqrt{a}\cdot\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\cdot\left[1-\frac{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)
\(=\left(1+\sqrt{a}\right)\cdot\left(1-\sqrt{a}\right)\)
= 1 - a
a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\) b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)
c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)
d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)
c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)
\(M=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\left(1+\dfrac{1}{a}\right)\)
\(M=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right)\left(1+\dfrac{1}{a}\right)\)
\(M=\left(\dfrac{\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{2}{2\left(1-a\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{1}{1-a}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\left(\dfrac{1+a-a^2-1}{\left(1-a\right)\left(1+a\right)}\right)\left(\dfrac{a+1}{a}\right)\)
\(M=\dfrac{a-a^2}{a\left(1-a\right)}\)
\(M=\dfrac{a\left(1-a\right)}{a\left(1-a\right)}=1\)
--> giá trị của M ko phụ thuộc vào A
a,\(\frac{x}{\sqrt{x}+1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}-1\right)+\frac{1}{\sqrt{x}-1}+2\ge2.\sqrt{\left(\sqrt{x}-1\right).\frac{1}{\sqrt{x}-1}+2}\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow\sqrt{x}-1=1\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\left(t/m\right)\)
Dmin = 4 <=> x=4
b,\(\frac{\sqrt{x-9}}{5x}\)
\(\sqrt{x-9}=\sqrt{\frac{\left(x-9\right).9}{9}}=\frac{1}{3}.\sqrt{\left(x-9\right).9}\le\frac{1}{3}.\frac{x-9+9}{2}=\frac{x}{2}\)
\(\Rightarrow D\le\frac{x}{\frac{6}{5x}}=\frac{x}{30x}=\frac{1}{30}\)
Dấu "=" xảy ra \(\Leftrightarrow x-9=9\Leftrightarrow x=18\)
Dmax=\(\frac{1}{30}\Leftrightarrow x=18\)
P/s : ko chắc lắm
\(a)\)\(P=\frac{x}{\sqrt{x}+1}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(P=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{1}{\sqrt{x}+1}}-2=2-2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}+1=\frac{1}{\sqrt{x}+1}\)\(\Leftrightarrow\)\(x=0\)
...
B= \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)= \(\frac{1}{20}\)
vậy B= \(\frac{1}{20}\)
Quy đồng lên rồi rút gọn là được