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Bài làm:
a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)
\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)
\(A=4+2\sqrt{3}+5\sqrt{3}-1\)
\(A=3+7\sqrt{3}\)
b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)
1) Ta có: \(5\left(x-2\right)=3x+10\)
\(\Leftrightarrow5x-10-3x-10=0\)
\(\Leftrightarrow2x-20=0\)
\(\Leftrightarrow2\left(x-10\right)=0\)
Vì 2>0
nên x-10=0
hay x=10
Vậy: x=10
2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy: x∈{-2;2;5}
3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)
\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)
\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)
\(\Leftrightarrow-5x=0\)
hay x=0
Vậy: x=0
4) ĐKXĐ: x≠5; x≠-5
Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow9x+45+14x-70-180=0\)
\(\Leftrightarrow23x-205=0\)
\(\Leftrightarrow23x=205\)
hay \(x=\frac{205}{23}\)(tm)
Vậy: \(x=\frac{205}{23}\)
\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}+\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}+\frac{x-1}{x+3}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\div\frac{x+2}{x+3}\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-x-12}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\left(\frac{21+x^2-x-12+x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x+12}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Đến đây là chịu ạ :(
a, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\frac{8x-3}{4}-\frac{6x-4}{4}=\frac{4x-2}{4}+\frac{x+3}{4}\)
\(8x-3-6x-4=4x-2+x+2\)
\(2x-7=5x\Leftrightarrow2x-5x+7=0\Leftrightarrow-3x=-7\Leftrightarrow x=\frac{7}{3}\)
b, \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
\(\frac{3x+9}{6}-\frac{2x-2}{6}=\frac{x+5}{6}+\frac{6}{6}\)
\(3x+9-2x-2=x+5+6\)
\(x+7=x+11\Leftrightarrow-4\ne0\)
\(\frac{\left(2n+1\right)^3+n^3}{\left(n+1\right)^3-n^3}=\frac{\left(3n+1\right)\left(3n^2+3n+1\right)}{3n^2+3n+1}=3n+1\)
\(\Rightarrow A=\left(3.1+1\right)+\left(3.2+1\right)+...+\left(3.20+1\right)\)
\(=3\left(1+2+...+20\right)+20\)
\(=\frac{3.20.21}{2}+20=...\)