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5 tháng 12 2021
\(c,=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x-2}\\ d,=\dfrac{\left(2-x-3\right)\left(2+x+3\right)}{\left(x+5\right)^2}=\dfrac{\left(x+5\right)\left(-x-1\right)}{\left(x+5\right)^2}=\dfrac{-x-1}{x+5}\)
HH
4
15 tháng 2 2022
\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
25 tháng 10 2021
\(6,=\left(x-2y\right)\left(5x^2-15x\right)=5x\left(x-3\right)\left(x-2y\right)\left(C\right)\\ 7,\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\left(D\right)\\ 8,=x^2-20x+100+1=\left(x-10\right)^2+1>0\left(A\right)\\ 9,\widehat{D}=360^0-120^0-80^0-100^0=60^0\left(D\right)\\ 10,D\)
25 tháng 10 2021
câu 6;
\(5x^2\left(x-2y\right)-15x\left(x-3y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\)
câu 7:
\(x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Câu 8:
\(x^2-20x+101=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1>0\)
\(\Rightarrow x^2-20x+101\) luôn dương
Câu 9:
\(\widehat{D}=360^o-120^o-80^o-100^o=60^o\)
Câu 10:D
22 tháng 11 2023
5: \(\left(2x-3\right)^2-16\)
\(=\left(2x-3\right)^2-4^2\)
\(=\left(2x-3-4\right)\left(2x-3+4\right)\)
\(=\left(2x-7\right)\left(2x+1\right)\)
6: \(\left(2x+1\right)^2-\left(\dfrac{1}{2}x-7\right)^2\)
\(=\left(2x+1-\dfrac{1}{2}x+7\right)\left(2x+1+\dfrac{1}{2}x-7\right)\)
\(=\left(\dfrac{3}{2}x+8\right)\left(\dfrac{5}{2}x-6\right)\)
7: \(\left(x+7\right)^2-\left(4x+5\right)^2\)
\(=\left(x+7-4x-5\right)\left(x+7+4x+5\right)\)
\(=\left(-3x+2\right)\left(5x+12\right)\)
8: \(\left(\dfrac{1}{4}x+3\right)^2-\left(4x-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{4}x+3-4x+\dfrac{1}{4}\right)\left(\dfrac{1}{4}x+3+4x-\dfrac{1}{4}\right)\)
\(=\left(-\dfrac{15}{4}x+\dfrac{13}{4}\right)\left(\dfrac{17}{4}x+\dfrac{11}{4}\right)\)
9: \(\left(\dfrac{1}{3}x+3\right)^2-\left(4x-\dfrac{2}{3}\right)^2\)
\(=\left(\dfrac{1}{3}x+3-4x+\dfrac{2}{3}\right)\left(\dfrac{1}{3}x+3+4x-\dfrac{2}{3}\right)\)
\(=\left(-\dfrac{11}{3}x+\dfrac{11}{3}\right)\left(\dfrac{13}{3}x+\dfrac{7}{3}\right)\)
\(=-\dfrac{11}{3}\left(x-1\right)\left(\dfrac{13}{3}x+\dfrac{7}{3}\right)\)
10: \(9\left(x+5\right)^2-4\left(2x+8\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left[2\left(2x+8\right)\right]^2\)
\(=\left(3x+15\right)^2-\left(4x+16\right)^2\)
\(=\left(3x+15-4x-16\right)\left(3x+15+4x+16\right)\)
\(=\left(-x-1\right)\left(7x+31\right)\)
11: \(\left(x+5\right)^2-25\left(2x+8\right)^2\)
\(=\left(x+5\right)^2-\left[5\left(2x+8\right)\right]^2\)
\(=\left(x+5\right)^2-\left(10x+40\right)^2\)
\(=\left(x+5-10x-40\right)\left(x+5+10x+40\right)\)
\(=\left(-9x-35\right)\left(11x+45\right)\)
12: \(16\left(2x+5\right)^2-9\left(2x-1\right)^2\)
\(=\left[4\left(2x+5\right)\right]^2-\left[3\left(2x-1\right)\right]^2\)
\(=\left(8x+20\right)^2-\left(6x-3\right)^2\)
\(=\left(8x+20+6x-3\right)\left(8x+20-6x+3\right)\)
\(=\left(14x+17\right)\left(2x+23\right)\)
17 tháng 1 2022
1: \(=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x\cdot2y=4xy\)
2: \(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
3: \(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
\(=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)
4: \(=\left(7-4x+5\right)\left(7+4x-5\right)=\left(12-4x\right)\left(4x+2\right)\)
\(=8\left(3-x\right)\left(2x+1\right)\)
a: ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+8x^3+1-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2-54=0\)
\(\Leftrightarrow9x^3+6x^2+27x-26-9x^3-6x^2-x=0\)
\(\Leftrightarrow26x=26\)
hay x=1
b: Ta có: \(\left(x-3\right)^3-\left(x^3-27\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2+2x+1\right)+3x^2+33=0\)
\(\Leftrightarrow-6x^2+27x+33+6x^2+12x+6=0\)
\(\Leftrightarrow39x=-39\)
hay x=-1