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PN
17 tháng 6 2020
\(M=\frac{2\sqrt{a}\left(\sqrt{a}+\sqrt{2a}-\sqrt{3b}\right)+\sqrt{3b}\left(2\sqrt{a}-\sqrt{3b}\right)-2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\left(đkxđ:a,b\ge0;mau\ne0\right)\)[tự tìm cái sau :)) ]
\(VP=\frac{2\sqrt{a}\left(\sqrt{a}+\sqrt{2}.\sqrt{a}-\sqrt{3}.\sqrt{b}\right)}{a\sqrt{2}+\sqrt{3ab}}+\frac{\sqrt{3b}\left(2\sqrt{a}-\sqrt{3b}\right)}{a\sqrt{2}+\sqrt{3ab}}-\frac{2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\)
\(=\frac{2a+2a\sqrt{2}-2\sqrt{3ab}}{a\sqrt{2}+\sqrt{3ab}}+\frac{2\sqrt{3ab}-3b}{a\sqrt{2}+\sqrt{3ab}}-\frac{2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\)
\(=\frac{2a+2a\sqrt{2}-3b+2a\sqrt{a}}{a\sqrt{2}+\sqrt{3ab}}\)
mình làm được đến đây , bạn làm được tiếp thì làm =))
NL
2
17 tháng 9 2018
\(\dfrac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}=\dfrac{2a-2\sqrt{ab}+3\sqrt{ab}-3b}{2a-2\sqrt{ab}-3\sqrt{ab}+3b}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}+3\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}-3\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+3\sqrt{b}}{2\sqrt{a}-3\sqrt{b}}\)
17 tháng 9 2018
\(ĐK:a,b\ge0;a\ne b.\dfrac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}+3\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}-3\sqrt{b}\right)}=\dfrac{2\sqrt{a}+3\sqrt{b}}{2\sqrt{a}-3\sqrt{b}}\)
10 tháng 8 2015
Điều kiện: x \(\ne\) 1; 1/4 ; x \(\ge\) 0
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\left(2a+\sqrt{a}-1\right).\sqrt{a}}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)-\left(2a+\sqrt{a}-1\right)\left(1+\sqrt{a}\right).\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(a+\sqrt{a}+1-a-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{2\sqrt{a}-1}\right)=1+\frac{-\sqrt{a}}{a+\sqrt{a}+1}=\frac{a+1}{a+\sqrt{a}+1}\)
Các bài tập dạng này hoàn toàn làm tương tự!!!
AH
Akai Haruma
Giáo viên
24 tháng 2 2020
Bài 1:
$14+\sqrt{40}+\sqrt{56}+\sqrt{140}=14+\sqrt{56}+(\sqrt{40}+\sqrt{140})$
=14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}=(12+2\sqrt{35})+2+(2\sqrt{10}+2\sqrt{14})$
$=(\sqrt{5}+\sqrt{7})^2+2+2\sqrt{2}(\sqrt{5}+\sqrt{7})$
$=(\sqrt{5}+\sqrt{7}+\sqrt{2})^2$
$\Rightarrow \sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}=\sqrt{2}+\sqrt{5}+\sqrt{7}$
\(\Rightarrow A=\frac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=1\)
AH
Akai Haruma
Giáo viên
24 tháng 2 2020
Lời giải:
a) ĐKXĐ: $a,b\geq 0$ và $a,b$ không đồng thời cùng bằng $0$
\(B=\frac{2a+2\sqrt{2}a-2\sqrt{3ab}+2\sqrt{3ab}-3b-2a\sqrt{2}}{a\sqrt{2}+\sqrt{3ab}}=\frac{2a-3b}{\sqrt{a}(\sqrt{2a}+\sqrt{3b})}=\frac{(\sqrt{2a}-\sqrt{3b})(\sqrt{2a}+\sqrt{3b})}{\sqrt{a}(\sqrt{2a}+\sqrt{3b})}\)
\(=\frac{\sqrt{2a}-\sqrt{3b}}{\sqrt{a}}=\sqrt{2}-\sqrt{\frac{3b}{a}}\)
b)
\(a=1+3\sqrt{2}; 3b=30+11\sqrt{8}\Rightarrow \frac{3b}{a}=\frac{30+11\sqrt{8}}{1+3\sqrt{2}}=\frac{(30+11\sqrt{8})(1-3\sqrt{2})}{(1+3\sqrt{2})(1-3\sqrt{2})}\)
\(=\frac{102+68\sqrt{2}}{17}=6+4\sqrt{2}=(2+\sqrt{2})^2\)
\(\Rightarrow \sqrt{\frac{3b}{a}}=2+\sqrt{2}\)
\(\Rightarrow B=\sqrt{2}-(2+\sqrt{2})=-2\)
29 tháng 6 2019
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{\sqrt{a^3}-3a\sqrt{b}+3\sqrt{a}.b-\sqrt{b^3}+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{3\sqrt{a^3}-3a\sqrt{b}+3b\sqrt{a}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
28 tháng 6 2019
\(S=\frac{\left[\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(S=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^2\sqrt{a}+\left(\sqrt{b}\right)^2\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{b}-\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(S=\frac{\left(\sqrt{a}\right)^3-3\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{3\left(\sqrt{a}\right)^3-3a\sqrt{b}+3\sqrt{a}b}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\sqrt{a}\left[\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3\right]}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
26 tháng 10 2020
Mong mọi người giúp đỡ mình , mình đang cần gấp , cảm ơn mọi người
26 tháng 10 2020
Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)
\(P=\left(\frac{2a+1}{a\sqrt{a}-1}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\times\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\)
ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(=\left(\frac{2a+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\times\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(\frac{2a+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\times\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(\frac{2a+1-a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\times\left(1-\sqrt{a}+a-\sqrt{a}\right)\)
\(=\frac{a+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\times\left(a-2\sqrt{a}+1\right)\)
\(=\frac{1}{\sqrt{a}-1}\times\left(\sqrt{a}-1\right)^2\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\)
b) \(P\times\sqrt{1-a}\)
\(=\left(\sqrt{a}-1\right)\times\sqrt{1-a}\)
ĐKXĐ: \(0\le x< 1\)
Với \(0\le x< 1\)
Ta có :\(\hept{\begin{cases}\sqrt{a}\le\sqrt{1}=1\Rightarrow\sqrt{a}-1\le0\\\sqrt{1-a}\ge0\end{cases}}\)
\(\Rightarrow\left(\sqrt{a}-1\right)\left(\sqrt{1-a}\right)\le0\)
\(=\frac{2a-2\sqrt{ab}+3\sqrt{ab}-3b}{2a-2\sqrt{ab}-3\sqrt{ab}+3b}=\frac{2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(2\sqrt{a}-3\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{a}+3\sqrt{b}}{2\sqrt{a}-3\sqrt{b}}\)