a: \(\sqrt{252}+\dfrac{1}{3}\sqrt{63}-\sqrt{175}\)

\(=4\sqrt{7}+\sqrt{7}-5\sqrt{7}\)

=0

a, bạn chụp rõ hơn đc ko ? 

b, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)đk : \(x\ge3;x\le-3\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

TH1 : x = 3

TH2 : \(\sqrt{x+3}=3\Leftrightarrow x+3=9\Leftrightarrow x=6\)

bổ sung c 

\(D=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)

\(\sqrt{2}D=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{5}-1-\sqrt{5}-1=-2\)

\(D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)

e: \(\Leftrightarrow\left\{{}\begin{matrix}10y=40\\4x+7y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\4x=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=-3\end{matrix}\right.\)

\(d,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2y+11}{3}\\\dfrac{4\left(2y+11\right)}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2y+11}{3}\\8y+44-15y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2y+11}{3}\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10+11}{3}=7\\y=5\end{matrix}\right.\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}4x=16-7y\\16-7y-3y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=16-7y\\10y=40\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x=16-7\cdot4=-12\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=4\end{matrix}\right.\)

\(g,\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\x-15y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=15y+16\\30y+32+10y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=15y+8\\40y=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{33}{40}\cdot15+8=\dfrac{29}{8}\\y=-\dfrac{33}{40}\end{matrix}\right.\)

\(h,\Leftrightarrow\left(x-15\right)\left(y+2\right)=\left(x+15\right)\left(y-1\right)\\ \Leftrightarrow xy+2x-15y-30=xy-x+15y-15\\ \Leftrightarrow3x=30y+15\Leftrightarrow x=10y+5\\ \text{Ta có }PT\left(1\right)\Leftrightarrow xy+2x-15y-30=xy\Leftrightarrow2x-15y=30\\ \Leftrightarrow2\left(10y+5\right)-15y=30\\ \Leftrightarrow y=4\Leftrightarrow x=45\)

\(i,\Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\\sqrt{5}\left(5\sqrt{5}-2y\right)+y=5+2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\25-2\sqrt{5}y+y=5+2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\y\left(1-2\sqrt{5}\right)=20-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\y=\dfrac{20-2\sqrt{5}}{1-2\sqrt{5}}=\dfrac{2\sqrt{5}\left(2\sqrt{5}-1\right)}{-\left(2\sqrt{5}-1\right)}=-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}+4\sqrt{5}=9\sqrt{5}\\y=-2\sqrt{5}\end{matrix}\right.\)

Câu i sửa: \(PT\left(2\right)\Leftrightarrow y=5+2\sqrt{5}-\sqrt{5}x\)

Thế vào \(PT\left(1\right)\Leftrightarrow x+10+4\sqrt{5}-2\sqrt{5}x=5\sqrt{5}\)

\(\Leftrightarrow x\left(1-2\sqrt{5}\right)=\sqrt{5}-10\\ \Leftrightarrow x=\dfrac{\sqrt{5}-10}{1-2\sqrt{5}}=\dfrac{\sqrt{5}\left(1-2\sqrt{5}\right)}{1-2\sqrt{5}}=\sqrt{5}\\ \Leftrightarrow y=5+2\sqrt{5}-5=2\sqrt{5}\)

\(\left(a-\dfrac{a^2+b^2}{a+b}\right)\left(\dfrac{1}{b}+\dfrac{2}{a-b}\right)\)

\(=\dfrac{a^2+ab-a^2-b^2}{a+b}\cdot\dfrac{a-b+2b}{b\left(a-b\right)}\)

\(=\dfrac{b\left(a-b\right)}{a+b}\cdot\dfrac{a+b}{b\left(a-b\right)}\)

=1

\(\left(a-\dfrac{x^2+a^2}{x+a}\right)\left(\dfrac{2a}{x}-\dfrac{4a}{x-a}\right)\)

\(=\dfrac{ax+a^2-x^2-a^2}{x+a}\cdot\dfrac{2ax-2a^2-4ax}{x\left(x-a\right)}\)

\(=\dfrac{a\left(x-a\right)}{x+a}\cdot\dfrac{-2a^2-2ax}{x\left(x-a\right)}\)

\(=\dfrac{a}{x+a}\cdot\dfrac{-2a\left(a+x\right)}{x}\)

\(=\dfrac{-2a^2}{x}\)

\(\left(a-\dfrac{x^2+a^2}{x+a}\right)\left(\dfrac{2a}{x}-\dfrac{4a}{x-a}\right)\)

\(=\dfrac{ax+a^2-x^2-a^2}{x+a}.\dfrac{2a\left(x-a\right)-4ax}{x\left(x-a\right)}\)

\(=\dfrac{x\left(a-x\right)}{x+a}.\dfrac{-2a^2-2ax}{x\left(x-a\right)}\)

\(=\dfrac{x\left(a-x\right)}{x+a}.\dfrac{-2a\left(a+x\right)}{x\left(x-a\right)}\)

\(=2a\)

Ta có: \(D=2x^2-8x-10\)

\(=2\left(x^2-4x-5\right)\)

\(=2\left(x^2-4x+4-9\right)\)

\(=2\left(x-2\right)^2-18\ge-18\forall x\)

Dấu '=' xảy ra khi x=2

\(D=2x^2-8x-10=2\left(x^2-4x+4\right)-8-10=2\left(x-2\right)^2-18\)

Do \(2\left(x-2\right)^2\ge0\)

\(\Rightarrow D=2\left(x-2\right)^2-18\ge-18\)

\(minD=-18\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

a,\(3xy+6x^2=3x\left(y+2x\right)\)

b,\(-4xy^2+8x^2y=-4xy\left(y-2x\right)\)

c,\(4\left(x+y\right)+x^2+xy=4\left(x+y\right)+x\left(x+y\right)\)\(=\left(4+x\right)\left(x+y\right)\)

d,\(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)

e,\(25x^2-\dfrac{1}{4}=\left(5x\right)^2-\left(\dfrac{1}{2}\right)^2=\left(5x-\dfrac{1}{2}\right)\left(5x+\dfrac{1}{2}\right)\)

a. \(=3x\left(y+2x\right)\)

b. \(=4xy\left(-y+2x\right)\)

c. \(=4\left(x+y\right)+x\left(x+y\right)=\left(4+x\right)\left(x+y\right)\)

d. \(=\left(2x+3y\right)\left(2x-3y\right)\)

e. \(=\left(5x+\dfrac{1}{2}\right)\left(5x-\dfrac{1}{2}\right)\)

f. \(=\left(4x+4+3x-6\right)\left(4x+4-3x+6\right)=\left(7x-2\right)\left(x+10\right)\)

g . \(=x^2+x+9x+9=x\left(x+1\right)+9\left(x+1\right)=\left(x+9\right)\left(x+1\right)\)