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0C
4
R
1
19 tháng 6 2021
\(5,A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(A=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(A=\left|2x-1\right|+\left|2x-3\right|\)
\(A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)
\(A\ge2\)
\(< =>MIN:A=2\)dấu = xảy khi \(\frac{1}{2}\le x\le\frac{3}{2}\)
LN
4 tháng 6 2021
idcm888dkk8cdw6ysgyxdbwdqjhqwuiowqqwudcgqofyhrli2uiy3yuyewiohewuwfwou
31 tháng 10 2020
SUy ra 2 trường hợp => từ 1 và 2 suy ra gì gì đó........
CHúc bạn hok tốt ;-;
TA
31 tháng 10 2020
Áp dụng căn bậc hai,ta từ 1 có thể suy ra 2(2 ở đây là 2TH).Ví dụ:
\(1=\sqrt{1}=\hept{\begin{cases}-1\\1\end{cases}}\)
Còn nếu từ số một suy ra số 2 thì :
\(2-2+1\)
\(=2-\left(1+1\right)+\left(0,5+0,5\right)\)
\(=2-\left(1+\sqrt{1}\right)+\left(0,5+\sqrt{0,25}\right)\)
\(=2-\left(1+-1\right)+\left(0,5+-0,5\right)\)
\(=2-\left(1-1\right)+\left(0,5-0,5\right)\)
\(=2-0+0\)
\(=2\)
HG
giải giúp câu c d bài 7 đáp án bị thầy che rồi
2
3 tháng 7 2021
\(7:a,\sqrt{2-x}=3\)
\(\left|2-x\right|=3^2=9\)
\(\orbr{\begin{cases}2-x=9\\2-x=-9\end{cases}\orbr{\begin{cases}x=-7\left(KTM\right)\\x=11\left(TM\right)\end{cases}}}\)
\(b,\sqrt{4-4x+x^2}=3\)
\(\sqrt{\left(2-x\right)^2}=3\)
\(\left|2-x\right|=3\)
\(\orbr{\begin{cases}2-x=3\\2-x=-3\end{cases}\orbr{\begin{cases}x=-1\left(TM\right)\\x=5\left(TM\right)\end{cases}}}\)
\(c,\sqrt{4+x^2}+x=3\)
\(\sqrt{4+x^2}=3-x\)
\(4+x^2=\left(3-x\right)^2\)
\(4+x^2=9-6x+x^2\)
\(x=\frac{5}{6}\left(TM\right)\)
\(d,\frac{1}{2}\sqrt{16x-32}-2\sqrt{4x-8}+\sqrt{9x-18}=5\)
\(2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=5\)
\(\sqrt{x-2}\left(2-4+3\right)=5\)
\(\sqrt{x-2}=5\)
\(\left|x-2\right|=25\)
\(\orbr{\begin{cases}x-2=25\\x-2=-25\end{cases}\orbr{\begin{cases}x=27\left(TM\right)\\x=-23\left(KTM\right)\end{cases}}}\)
HG
giải giúp em bài 14 15 16 17 của bài 14 nha
2
26 tháng 6 2021
14, \(\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)
\(=\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}\)
\(=\frac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(3\sqrt{x}+14\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}+14}{5\sqrt{x}-1}\)
1) \(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b}{a-b}=\frac{4\sqrt{ab}}{a-b}\)
2) \(x-4-\sqrt{16-8x^2+x^4}=x-4-\sqrt{\left(x^2-4\right)^2}=x-4-\left|x^2-4\right|\)
3) \(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a\)
4) \(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=a-b\)
5) \(\frac{a-3\sqrt{a}}{\sqrt{a}-3}-\frac{a+4\sqrt{a}+3}{\sqrt{a}+3}=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}-\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\)
\(=\sqrt{a}-\sqrt{a}-1=-1\)
6) \(\frac{9-x}{\sqrt{x}+3}-\frac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6=\frac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=3-\sqrt{x}-\sqrt{x}+3-6=-2\sqrt{x}\)
7) \(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{\sqrt{x}-\sqrt{y}}{x-y}=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}:\frac{\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2\)
8) \(\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
=\(\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)
9) \(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
10) \(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}+\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
\(=\left(\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\left(\sqrt{x}-3\right)^2}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{2-\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4x+8\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{4\sqrt{x}}{\sqrt{x}-3}\)
11) \(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{1}{\sqrt{x}+2}\)
12) \(\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}+1\right)\left(\frac{\sqrt{x}-x}{\sqrt{x}-1}+1\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right)\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right)\)
\(=\left(\sqrt{x}+1\right)^2\)