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5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
2)
\(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}\)
\(=\sqrt{12,1.36.10}\)
= \(\sqrt{121.36}\)
\(=\sqrt{4356}\)
\(=66\)
3)
\(\sqrt{5a}.\sqrt{45a}-3a\)
\(=\sqrt{5.45a^2}-3a\)
\(=\sqrt{225a^2}-3a\)
\(=\sqrt{\left(15a\right)^2}-3a\)
\(=-15a-3a\) ( vì \(a\le0\))
\(=-18a\)
5)
\(\sqrt{0,36a^2}\)
\(=\sqrt{\left(0,6a\right)^2}\)
\(=-0,6a\) ( vì \(a< 0\) )
Để tối mình rảnh lên coi có làm tiếp được nữa hông thì mình làm ha.
Chúc bạn học tốt!
1)
\(\sqrt{3a^3}.\sqrt{12}\)
\(=\sqrt{3}.\sqrt{a^3}.\sqrt{12}\)
\(=\sqrt{3.12}.\sqrt{a^3}\)
\(=6\sqrt{a^3}\)
4)
\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=9.6a.a^2-\sqrt{0,2}.\sqrt{18}.\sqrt{10}.\sqrt{a^2}\)
\(=54a^3-\sqrt{2}.\sqrt{18}.\sqrt{a^2}\)
\(=34a^3-\sqrt{2.18}.\sqrt{a^2}\)
\(=54a^3-6\sqrt{a^2}\)
\(=54a^3-6a^2\) ( vì a<0)
6)
\(\sqrt{a^4.\left(3-a^{ }\right)^2}\)
\(=\sqrt{\left(a^2\right)^2.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\right)^2}.\sqrt{\left(3-a\right)^2}\)
\(=\left|a^2\right|\left|3-a\right|\) ( vì a>3 => a>3 nên 3-a<0)
Mà\(\left|3-a\right|=-\left(-3-a\right)=-3+a=a-3\)
\(=a^2\left(a-3\right)\)
\(=a^3-3a^2\)
Còn lại bạn làm tương tự nha, trể quá rùi :)))))
a) \(A=\sqrt{81}.\sqrt{\frac{9}{4}}+2\sqrt{16}-3=\sqrt{9^2}.\sqrt{\left(\frac{3}{2}\right)^2}+2\sqrt{4^2}-3=9.\frac{3}{2}+2.4-3=\frac{37}{2}\)
b) \(B=\sqrt{9-2\sqrt{14}}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\sqrt{7}-\sqrt{2}\)
c) Không rút gọn được.
Bài 2 : Mình hướng dẫn thôi nhé ^^
a) \(M=x^2-10x+30=\left(x^2-10x+25\right)+5=\left(x-5\right)^2+5\ge5\)
b) \(N=4x^2-12x+1=\left[\left(2x\right)^2-12x+9\right]-8=\left(2x-3\right)^2-8\ge-8\)
c) \(P=x^2-x-1=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
d) \(Q=16x^2-8x+3=\left[\left(4x\right)^2-8x+1\right]+2=\left(4x-1\right)^2+2\ge2\)
e) \(H=\frac{1}{9}x^2+3x-1=\left[\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{9}{2}+\frac{81}{4}\right]-\frac{81}{4}-1=\left(\frac{x}{3}+\frac{9}{2}\right)^2-\frac{85}{4}\ge-\frac{85}{4}\)
c, ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}-1=2\\\sqrt{2x-1}-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{2x-1}=-1\left(vn\right)\end{matrix}\right.\)
\(\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tm\right)\)
a, ĐKXĐ: \(x\in R\)
\(\sqrt{3x^2}=x+2\)
\(\Leftrightarrow\sqrt{3}\left|x\right|=x+2\)
TH1: \(\sqrt{3}x=x+2\)
\(\Leftrightarrow\left(\sqrt{3}-1\right)x=2\)
\(\Leftrightarrow x=\sqrt{3}+1\)
TH2: \(\sqrt{3}x=-x-2\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)x=-2\)
\(\Leftrightarrow x=1-\sqrt{3}\)
\(a\sqrt{b}-b\sqrt{a}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(7\sqrt{7}+3\sqrt{3}=\left(\sqrt{7}+\sqrt{3}\right)\left(7-\sqrt{21}+3\right)=\left(\sqrt{7}+\sqrt{3}\right)\left(10-\sqrt{21}\right)\)
\(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)
\(1-a\sqrt{a}=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)\)
\(x^2-\sqrt{x}=\sqrt{x}\left(x\sqrt{x}-1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(\left(\sqrt{2}+1\right)^2-4\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\left(\sqrt{5}+2\right)^2-8\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
2 cái trên đều áp dụng HĐT \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
\(5\sqrt{2}-2\sqrt{5}=\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)\)