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em tiểu học nên chẳng biết mà ít có người lớp 9 như chị
Bài 1:
b) Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\frac{\sqrt{2\left(4+\sqrt{7}\right)}}{\sqrt{2}}-\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}-\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\frac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Bài 2:
a) Ta có: \(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)\)
\(=a-\sqrt{a}-a-\sqrt{a}\)
\(=-2\sqrt{a}\)
b) Ta có: \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{ab}-\sqrt{ab}=0\)
d) Ta có: \(\frac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\sqrt{a}+\sqrt{b}-\left(\sqrt{a}+\sqrt{b}\right)\)
=0
Bài 3:
a) ĐKXĐ: x≥0
Ta có: \(\frac{\sqrt{27x}}{\sqrt{3}}=6\)
\(\Leftrightarrow\frac{\sqrt{27}\cdot\sqrt{x}}{\sqrt{3}}=6\)
\(\Leftrightarrow3\cdot\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{3}=2\)
hay \(x=4\)(thỏa mãn)
Vậy: S={4}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-1\end{matrix}\right.\Leftrightarrow x\ge0\)
Ta có: \(\sqrt{x+1}=3-\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=\left(3-\sqrt{x}\right)^2\)
\(\Leftrightarrow x+1=9-6\sqrt{x}+x\)
\(\Leftrightarrow x+1-9+6\sqrt{x}-x=0\)
\(\Leftrightarrow-8+6\sqrt{x}=0\)
\(\Leftrightarrow6\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=\frac{8}{6}=\frac{4}{3}\)
hay \(x=\frac{16}{9}\)(thỏa mãn)
Vậy: \(S=\left\{\frac{16}{9}\right\}\)
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
3/a) \(BĐT\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)(đúng với mọi x, y không âm)
Đẳng thức xảy ra khi x = y
b) \(BĐT\Leftrightarrow\frac{\left(x-y\right)^2}{xy}\ge0\) (đúng với mọi x, y không âm)
"=" <=> x = y
c) BĐT \(\Leftrightarrow2a+2b+2\ge2\sqrt{ab}+2\sqrt{a}+2\sqrt{b}\)
\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(a-2\sqrt{a}+1\right)+\left(b-2\sqrt{b}+1\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{a}-1\right)^2+\left(\sqrt{b}-1\right)^2\ge0\) (đúng)
"=" <=> a = b = 1
1/ \(A=\sqrt{7-2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}-\sqrt{2}\right|\) (thực ra em nghĩ ko cần thêm trị tuyệt đối đâu nhưng thêm cho chắc:D)
\(=\sqrt{7}-1-\sqrt{7}+\sqrt{2}=\sqrt{2}-1\)
2/Em thấy nó sai sai nên thôi:(
A= \(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{\sqrt{b}}{\sqrt{a}.\sqrt{a}-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-\sqrt{b}.\sqrt{b}}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{a}.\sqrt{a}.\sqrt{b}-\sqrt{b}.\sqrt{b}\sqrt{a}\right)\)
A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
A = b-a
B = \(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}}{a-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-1}\)
B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}\left(a+\sqrt{a}\right)}{a^2-a}\right).\frac{a-1}{\sqrt{a}+1}\)
B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}.\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
\(B=\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}-\frac{a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B= \(\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)-a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B= \(\left(\frac{\left(\sqrt{a}+1\right)\left(a\sqrt{a}-a\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B = \(\frac{\left(\sqrt{a}+1\right)a\left(\sqrt{a}-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(B=\frac{a\left(\sqrt{a}^2-1^2\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(B=\frac{a\left(a-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
B = \(\frac{a-1}{\sqrt{a}+1}\)
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
tính
\(\frac{a-\sqrt{ab}}{b-\sqrt{ab}}+\frac{b-\sqrt{ab}}{a+\sqrt{ab}}=\frac{a-ab+b-ab}{ab+b\sqrt{ab}-a\sqrt{ab}-ab}=\frac{a+b}{\sqrt{ab}\left(b-a\right)}\)
còn lại mk chịu
bạn ghi rõ hơn nữa được không chứ mình chưa hiểu lắm