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giải hộ mk với ạ phease!!! dễ nha ko chép mạng
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Giải thích các bước giải:
a)$\frac{√a+√b}{√a-√b}$+ $\frac{√a-√b}{√a+√b}$(a,b≥0,a$\neq$ b)
=$\frac{(√a+√b)²}{a-b}$+ $\frac{(√a-√b)²}{a-b}$
=$\frac{a+2√ab+b+a-2√ab+b}{a-b}$=$\frac{2a+2b}{a-b}$
=$\frac{2(a+b)}{a-b}$=2(a,b≥0,a$\neq$ b)
b)$\frac{a-b}{√a-√b}$-$\frac{√a³-√b³}{a-b}$(a,b≥0,a$\neq$ b)
=$\frac{(a-b)(√a+√b)}{a-b}$-$\frac{√a³-√b³}{a-b}$
=$\frac{√a³+a√b-b√a-√b³-√a³+√b³}{a-b}$
=$\frac{a√b-b√a}{a-b}$=$\frac{√ab}{√a+√b}$(a,b≥0,a$\neq$ b)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{\sqrt a + \sqrt b }}{{\sqrt a – \sqrt b }} + \dfrac{{\sqrt a – \sqrt b }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2} + {{\left( {\sqrt a – \sqrt b } \right)}^2}}}{{\left( {\sqrt a – \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{{\left( {a + 2\sqrt {ab} + b} \right) + \left( {a – 2\sqrt {ab} + b} \right)}}{{a – b}}\\
= \dfrac{{2.\left( {a + b} \right)}}{{a – b}}\\
b,\\
\dfrac{{a – b}}{{\sqrt a – \sqrt b }} – \dfrac{{\sqrt {{a^3}} – \sqrt {{b^3}} }}{{a – b}}\\
= \dfrac{{\left( {\sqrt a – \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a – \sqrt b }} – \dfrac{{\left( {\sqrt a – \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}{{\left( {\sqrt a – \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}\\
= \sqrt a + \sqrt b – \dfrac{{a + \sqrt {ab} + b}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2} – \left( {a + \sqrt {ab} + b} \right)}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\left( {a + 2\sqrt {ab} + b} \right) – \left( {a + \sqrt {ab} + b} \right)}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt {ab} }}{{\sqrt a + \sqrt b }}
\end{array}\)