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\(a\text{)}\:36x^2-5=\left(6x\right)^2-\left(\sqrt{5}\right)^2\\ =\left(6x-\sqrt{5}\right)\left(6x+\sqrt{5}\right)\)
\(b\text{)}\:25-3x^2=5^2-\left(\sqrt{3}x\right)^2\\ =\left(5-\sqrt{3}x\right)\left(5+\sqrt{3}\right)\)
\(c\text{)}\:x-4=\left(\sqrt{x}\right)^2-2^2\\ =\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(d\text{)}\:11+9x=9.\dfrac{11}{9}+9x\\ =9\left(\dfrac{11}{9}+x\right)\)
\(e\text{)}\:31+7x=7.\dfrac{31}{7}+7x\\ =7\left(\dfrac{31}{7}+x\right)\)
\(a,x-9+y-2\sqrt{xy}\left(x;y>0\right)\)
\(=\left(\sqrt{x}\right)^2-2\sqrt{x}\sqrt{y}+\left(\sqrt{y}\right)^2-9\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2-9\)
\(=\left(\sqrt{x}-\sqrt{y}+3\right)\left(\sqrt{x}-\sqrt{y}-3\right)\)
\(b,\text{ đkxđ }x\ge0\)
\(x-5\sqrt{x}+6=\left(\sqrt{x}\right)^2-2\sqrt{x}-3\sqrt{x}+6\)
\(=\sqrt{x}.\left(\sqrt{x}-2\right)-3.\left(\sqrt{x}-2\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)
\(c,đ\text{kxđ }x\ge0\)
\(x-2\sqrt{x}-3=\left(\sqrt{x}\right)^2+\sqrt{x}-3\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)+3.\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)
\(d,\text{đkxđ }x\ge0\)
\(\sqrt{x}-x^2=\sqrt{x}-\left(\sqrt{x}\right)^4=\sqrt{x}\left(1-\left(\sqrt{x}\right)^3\right)\)
\(=\sqrt{x}.\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)\)
câu a) rút x theo y thế vào A rồi áp dụng HĐT
b)rút xy thế vào B
c)HĐT
d)rút x theo y thé vào C
rồi dùng BĐT cô-si
e)BĐT chưa dấu giá trị tuyệt đối
a: \(=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)
b: \(=\left(x-\sqrt{2}\right)^2\)
c: \(=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
d: \(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)
e: \(=\left(x-\sqrt{23}\right)^2\)
a, \(x^2-49x-50=0\Leftrightarrow\left(x-1\right)\left(x+50\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-50\end{cases}}\)
b, \(3x^2-7x-10=0\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\Leftrightarrow\left(3x-10\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=10\\x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}}\)
c, \(x^2-4x-5=0\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
d, \(x^2+2x-3=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
e, \(x^2+2020x-2021=0\)
=> vô nghiệm
f, \(x^2+9x-10=0\Leftrightarrow\left(x-1\right)\left(x+10\right)\Leftrightarrow\orbr{\begin{cases}x=1\\x=-10\end{cases}}\)
g, \(-5x^2+4x+1=0\Leftrightarrow5x^2+x-5x-1=0\Leftrightarrow x\left(5x+1\right)-1\left(5x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{5}\end{cases}}\)
h, \(4x^2+3x-7=0\Leftrightarrow x\left(4x+7\right)-1\left(4x+7\right)=0\Leftrightarrow\left(x-1\right)\left(4x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{7}{4}\end{cases}}\)
a) (x-50)(x+1)=0
<=>x=50 hoặc x=1
b) (x+1)(x-10/3)=0
<=>x=-1 hoặc x=10/3
c) (x-5)(x+1)=0
<=>x=5 hoặc x=-1
d) (x+3)(x-1)=0
<=>x=-3 hoặc x=1
e) (x-1)(x+2021)=0
<=>x=1 hoặc x=-2021
f) (x-1)(x+10)=0
<=> x=1 hoặc x=-10
g) (x+1/5)(x-1)=0
<=>x=1 hoặc x=-1/5
h) (x-1)(x+7/4)=0
<=> x=1 hoặc x=-7/4
Học tốt. tk vs ạ
\(\sqrt{x^2\left(x-1\right)^2}=\left|x\left(x-1\right)\right|\)
\(x< 0\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x< 0\end{matrix}\right.\Leftrightarrow x\left(x-1\right)>0\Rightarrow\left|x\left(x-1\right)\right|=x\left(x-1\right)=x^2-x\)
\(b,\sqrt{13x}.\sqrt{\frac{52}{x}}=\sqrt{\frac{13.52.x}{x}}=\sqrt{13.52}=\sqrt{13^2.2^2}=\sqrt{26^2}=26\)
Lời giải :
a) \(\sqrt{x^2\left(x-1\right)^2}=\left|x\right|\cdot\left|x-1\right|=-x\left(1-x\right)=x^2-x\)
b) \(\sqrt{13x}\cdot\sqrt{\frac{52}{x}}=\sqrt{\frac{13x\cdot52}{x}}=\sqrt{676}=26\)
c) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)
d) \(\sqrt{\frac{9+12x+4x^2}{y^2}}=\sqrt{\frac{\left(2x+3\right)^2}{y^2}}=\frac{2x+3}{-y}=\frac{-2x-3}{y}\)
Bài làm:
a) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b) \(4x^2-5=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)
c) \(3x^2-1=\left(x\sqrt{3}-1\right)\left(x\sqrt{3}+1\right)\)
d) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
e) \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
f) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)