Đề bài là Rút gọn biểu thức nha . Mình quên ghi ^^

\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}=\sqrt{16.5}+\sqrt{9.5}+\sqrt{5}\)

\(=4\sqrt{5}+3\sqrt{5}+\sqrt{5}=8\sqrt{5}\)

\(B=\frac{5}{\sqrt{10}}+3,5\sqrt{40}=\sqrt{\frac{25}{10}}+3,5\sqrt{16.2,5}\)

\(=\sqrt{2,5}+3,5.4\sqrt{2,5}=15\sqrt{2,5}\)

\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)

\(=\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{100.3}}{10}-\sqrt{4.3}\)

\(=-\sqrt{3}-2+\sqrt{3}-2\sqrt{3}=-2\sqrt{3}-2\)

\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}=4\sqrt{x}+2x-4\sqrt{x}=2x\) ( do \(x\ge0\))

\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}.\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=\sqrt{a}\)

mk chỉnh đề

\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+4}\)

\(=\sqrt{25\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}\)

\(=5\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\)

\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}-\frac{2\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)

\(=\sqrt{3}-\sqrt{5}-\sqrt{5}-\sqrt{7}=\sqrt{3}-\sqrt{7}\)

\(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

\(=\left|3x-1\right|+\left|5-3x\right|\)

\(\ge\left|3x-1+5-3x\right|=4\)

a - b =3

a² - b² = 15

=> a+b = 5 

=> a =4 ; b = 1

=> 25 - x² = 16 => x = + -3 thỏa mãn 10 -x² =1

em moi hoc lop 7 thoi a

a)ĐKXĐ: \(x\ge0\)

Ta có: \(\sqrt{x}=2\)

\(\Leftrightarrow x=2^2=4\)(nhận)

Vậy: S={4}

b) ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{25x-25}-10=0\)

\(\Leftrightarrow\sqrt{25}\cdot\sqrt{x-1}=10\)

\(\Leftrightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=2^2=4\)

hay x=5(nhận)

Vậy: S={5}

c)ĐKXĐ: \(x\in Z\)

Ta có: \(\sqrt{25-10x+x^2}=7\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7\)

\(\Leftrightarrow\left|x-5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={12;-2}

\(a,\sqrt{x^4}=7\Leftrightarrow x^2=7\Leftrightarrow x=\pm\sqrt{7}\)

\(Dk:x\ge\frac{2}{3};\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\left(tm\right)\)

\(dk:x\ge\frac{3}{2};\sqrt{2x-3}=\sqrt{x-1}\Leftrightarrow2x-3=x-1\Leftrightarrow x=2\left(tm\right)\)

\(dk:x\ge0;x-10\sqrt{x}+25=0\Leftrightarrow\left(\sqrt{x}-5\right)^2=0\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)

\(\sqrt{2x}< 3\Leftrightarrow\sqrt{2}.\sqrt{x}< 3\Leftrightarrow0\le\sqrt{x}< \sqrt{4,5}\Leftrightarrow0\le x< 4,5\)

\(h,dk:x\ge3;\sqrt{\left(x-1\right)^2}=3x-9\Leftrightarrow\left|x-1\right|=3x-9\Leftrightarrow x-1=3x-9\left(x\ge3\right)\Leftrightarrow x=4\left(tm\right)\)