150 : ( x + 2 ) = 25
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^2+(x+1)^2=25
thử từ 1 đến 9
x=1=>1^2+(1+1)^2=5 (sai)
x=2=>2^2+(2+1)^2=13(sai)
x=3=>3^2+(3+1)^2=25(đúng)
Vậy X =3
\(x^2+\left(x+1\right)^2=25\)hoặc -25
Mà x^2 và (x+1)^2 đều là số dương
=> x^2 + (x+1)^2 = 25
Mà x^2 < 25, rồi bạn làm nốt nhé
Kết quả là x= -3 hoặc 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2M=\left(\sqrt{x^2-3x+25}-\sqrt{x^2-3x+9}\right)\)\(\left(\sqrt{x^2-3x+25}+\sqrt{x^2-3x+9}\right)\)
\(2M=x^2-3x+25-x^2+3x-9=16\)
M = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{x+5}-\sqrt{x}\right)\left(\sqrt{x+5}+\sqrt{x}\right)=\sqrt{x+5}+\sqrt{x}\)
=> \(x+5-x=M\Rightarrow M=5\)
b ) tương tự
b) N.N' = \(\left(\sqrt{25-x^2}-\sqrt{15-x^2}\right).\left(\sqrt{25-x^2}+\sqrt{15-x^2}\right)=\left(25-x^2\right)-\left(15-x^2\right)=10\)
=> 2.N = 10 => N = 10:2 =5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+4=13\\ \Rightarrow x^2=9\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ b,150-5\left(x-2\right)^2=25\\ \Rightarrow5\left(x-2\right)^2=125\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{3x-5}{x^2-5x}-\frac{x+5}{5x-25}\right):\frac{x^2-25}{x}\)
\(=\left[\frac{3x-5}{x\left(x-5\right)}-\frac{x+5}{5\left(x-5\right)}\right].\frac{x}{x^2-25}\)
\(=\left[\frac{\left(3x-5\right).5}{x\left(x-5\right).5}-\frac{\left(x+5\right).x}{5\left(x-5\right).x}\right].\frac{x}{x^2-25}\)
\(=\left[\frac{15x-25}{5x\left(x-5\right)}-\frac{x^2+5x}{5x\left(x-5\right)}\right].\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{15x-25-x^2-5x}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-x^2+10x-25}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-\left(x-5\right)^2.x}{5x\left(x-5\right)\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-1}{5\left(x+5\right)}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\sqrt{25x^2}=10\)
\(\sqrt{\left(5x\right)^2}=10\)
\(5x=10\)
\(x=2\)
b. <=> \(\sqrt{4\left(x^2-1\right)}=2\sqrt{15}\) ĐKXĐ: x>=1,x>=-1
<=> \(4\left(x^2-1\right)=60\Leftrightarrow x^2-1=15\Leftrightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)
<=>x=+-4
![](https://rs.olm.vn/images/avt/0.png?1311)
150 : ( x + 2 ) = 25
x + 2 = 150 : 25
x + 2 = 6
x = 6 - 2
x = 4
150 : (x + 2) = 25
x + 2 = 150 : 25
x + 2 = 6
x = 6 - 2
x = 4
@Bảo
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