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-Nếu \(x\le2\) thì A=-81-25(2-x)+40=-81-50+25x+40=25x-91
-Nếu \(x\ge2\) thì A=-81-25(x-2)+40=-81-25x+50+40=9-25x
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\sqrt{4a^2}=\sqrt{\left(2a\right)^2}=\left|2a\right|=2a\)
b/ \(\sqrt{\left(\frac{2}{5}\right)^2\left(x-2\right)^2}=\frac{2}{5}\left|x-2\right|=\frac{2}{5}\left(x-2\right)=\frac{2x}{5}-\frac{4}{5}\)
c/ \(\sqrt{5^2\left(3-a\right)^2}+3=5\left|3-a\right|+3=\left[{}\begin{matrix}18-5a\left(a\le3\right)\\5a-12\left(a\ge3\right)\end{matrix}\right.\)
d/ \(=\frac{1}{2\left(x-5\right)}.6\left|x-5\right|=\frac{3\left|x-5\right|}{x-5}=\left[{}\begin{matrix}3\left(x>5\right)\\-3\left(x< 5\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(F=\frac{5}{3-\sqrt{5}}-\frac{4}{3+\sqrt{5}}=\frac{5\left(3+\sqrt{5}\right)}{9-5}-\frac{4\left(3-\sqrt{5}\right)}{9-5}=\frac{15+5\sqrt{5}-12+4\sqrt{5}}{4}=\frac{3+9\sqrt{5}}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(2x^2+5x+3=2x^2+2x+3x+3=2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(2x+3\right)\)
\(\Rightarrow\left(\sqrt{2x+3}-\sqrt{x+1}\right)\left(\sqrt{2x^2+5x+3}+1\right)=x+2\left(ĐKXĐ:x\ge-1\right)\\ \Leftrightarrow\left(\sqrt{2x+3}-\sqrt{x+1}\right)\left(\sqrt{\left(2x+3\right)\left(x+1\right)}+1\right)=2x+3-\left(x+1\right)\left(1\right)\)
Đặt \(\sqrt{2x+3}=a\ge1,\sqrt{x+1}=b\ge0\), phương trình (1) trở thành:
\(\left(a-b\right)\left(ab+1\right)=a^2-b^2\)
\(\left(a-b\right)\left(ab+1\right)-\left(a-b\right)\left(a+b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left[a\left(b-1\right)-\left(b-1\right)\right]=0\\ \Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\\
\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
+) Với a=b ta có: \(\sqrt{2x+3}=\sqrt{x+1}\Leftrightarrow2x+3=x+1\Leftrightarrow x=-2\left(ktm\right)\)
+) Với a=1 ta có: \(\sqrt{2x+3}=1\Leftrightarrow2x+3=1\Leftrightarrow x=-1\left(tm\right)\)
+) Với b=1 ta có : \(\sqrt{x+1}=1\Leftrightarrow x+1=1\Leftrightarrow x=0\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-1;0\right\}\).
Tick cho mình nha <3 !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(pt\left(1\right)\Leftrightarrow x\left(x+2\right)+y\left(y+2\right)=11\)
Đặt a=x(x+2); b=y(y+2) thì: \(hpt\Leftrightarrow\hept{\begin{cases}a+b=11\\ab=24\end{cases}}\)
Khi đó a,b là 2 nghiệm của pt ẩn m:
\(m^2-11m+24=0\Leftrightarrow\left(m-8\right)\left(m-3\right)=0\Rightarrow\hept{\begin{cases}m=8\\m=3\end{cases}}\)
Tới đây bn tự làm tiếp.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)
\(=\sqrt{5^2\left(y+4\right)}+\sqrt{6^2\left(y+4\right)}-2\sqrt{9^2\left(y+4\right)}\)
\(=5\sqrt{y+4}+6\sqrt{y+4}-2\cdot9\sqrt{y+4}\)
\(=11\sqrt{y+4}-18\sqrt{y+4}\)
\(=-7\sqrt{y+4}\)
\(=5\sqrt{y+4}+6\sqrt{y+4}-2\cdot9\sqrt{y+4}\)
\(=11\sqrt{y+4}-18\sqrt{y+4}\)
\(=-7\sqrt{y+4}\)
A=-41-25.|2-x|
nếu 2-x>=0<=>x<=2
thì |2-x|=2-x
=>A=-41-25.(2-x)=-41-50+25x=25x-91
nếu 2-x<0<=>x>2
thì |2-x|=x-2
=>A=-41-25.(x-2)
=-41-25x+50
=9-25x
vậy A=25x-91 khi x<=2
A=9-25x khi x>2