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NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2021
a. ĐKXĐ:...
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}-3+3\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x}{2}-\dfrac{3}{x}\right)^2+6=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=t\Rightarrow2t^2-13t+6=0\Rightarrow\left[{}\begin{matrix}t=6\\t=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{3}{x}=6\\\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2021
b. ĐKXĐ: ...
\(\Leftrightarrow x\left(x-1\right)-\dfrac{x-1}{x^2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x=1\)
3 tháng 9 2022
Bài 2:
a: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x+1\right)^2=18\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=18\)
=>45x=9
=>x=1/5
b: \(\Leftrightarrow x^3-16x-x^3+125=13\)
=>-16x=-112
=>x=7
18 tháng 7 2017
a) Ta có : 6x(3x + 5) - 2x(9x - 2) + (17 - x)(x - 1) + x(x - 18) = 0
<=> 18x2 + 30x - 18x2 + 4x + 17x - 17 - x2 + x + x2 - 18x = 0
<=> 34x - 17 = 0
<=> 34x = 17
=> x = 2
1 tháng 11 2017
\(1,x^2+2xy+x+2y\)
\(=\left(x^2+2xy\right)+\left(x+2y\right)\)
\(=x\left(x+2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+1\right)\)
\(2,x^2-10x+25\)
\(=x^2-2.x.5+5^2\)
\(=\left(x-5\right)^2\)
Đợi mk chút ,mk có việc bận ,tối mk làm tiếp nha bn
1 tháng 11 2017
\(3,x^3+3x^2+3x+1\)
\(=\left(x^3+1\right)+\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+3x\right)\)
\(=\left(x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x+1\right)\left(x+1\right)^2\)
\(=\left(x+1\right)^3\)
\(4,x^3-8\)
\(=x^3-2^3\)
\(=\left(x-2\right)\left(x^2+2x+4\right)\)
\(5,x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
\(6,x^3-\dfrac{1}{8}\)
\(=x^3-\left(\dfrac{1}{2}\right)^3\)
\(=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
\(7,x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
\(8,4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
\(9,49x^2-9\)
\(=\left(7x\right)^2-3^2\)
\(=\left(7x-3\right)\left(7x+3\right)\)
BA
11 tháng 11 2016
a)x(x^2-x+2)(x^2+x+2)
còn lại thì vào đây: https://coccoc.com/search/math
5 tháng 9 2020
Áp dụng : (A + B)3 = A3 + 3A2B + 3AB2 + B3
11) \(\left(x^2+\frac{3}{xy}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{3}{xy}+3\cdot x^2\cdot\left(\frac{3}{xy}\right)^2+\left(\frac{3}{xy}\right)^3\)
\(=x^6+3\cdot x^4\cdot\frac{3}{xy}+3\cdot x^2\cdot\frac{9}{x^2y^2}+\frac{27}{x^3y^3}\)
\(=x^6+\frac{9x^4}{xy}+\frac{27\cdot x^2}{x^2y^2}+\frac{27}{x^3y^3}\)
\(=x^6+\frac{9x^3}{y}+\frac{27}{y^2}+\frac{27}{x^3y^3}\)
12) \(\left(x^2+\frac{2}{x}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{2}{x}+3\cdot x^2\cdot\left(\frac{2}{x}\right)^2+\left(\frac{2}{x}\right)^3\)
\(=x^6+3\cdot x^4\cdot\frac{2}{x}+3\cdot x^2\cdot\frac{4}{x^2}+\frac{8}{x^3}\)
\(=x^6+\frac{6\cdot x^4}{x}+\frac{12\cdot x^2}{x^2}+\frac{8}{x^3}\)
\(=x^6+6x^3+12+8x^3\)
13) \(\left(3y+\frac{x}{2}\right)^3=\left(3y\right)^3+3\cdot3y^2\cdot\frac{x}{2}+3\cdot3y+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3\)
\(=27y^3+\frac{9y^2\cdot x}{2}+9y+\frac{x^2}{4}+\frac{x^3}{8}\)
14) \(\left(1\frac{1}{2}xy+1\right)^3=\left(\frac{3}{2}xy+1\right)^3=\left(\frac{3}{2}xy\right)^3+3\cdot\left(\frac{3}{2}xy\right)^2\cdot1+3\cdot\frac{3}{2}xy\cdot1^2+1^3\)
\(=\frac{27}{8}x^3y^3+3\cdot\frac{9}{4}x^2y^2+\frac{9}{2}xy+1\)
\(=\frac{27}{8}x^3y^3+\frac{27}{4}x^2y^2+\frac{9}{2}xy+1\)
15) \(\left(\frac{x^2}{2}+\frac{2}{y}\right)^3=\left(\frac{x^2}{2}\right)^3+3\cdot\left(\frac{x^2}{2}\right)^2\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\left(\frac{2}{y}\right)^2+\left(\frac{2}{y}\right)^3\)
\(=\frac{x^6}{8}+3\cdot\frac{x^4}{4}\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\frac{4}{y^2}+\frac{8}{y^3}\)
\(=\frac{x^6}{8}+\frac{3x^4}{2y}+\frac{6x^2}{y^2}+\frac{8}{y^3}\)
Còn 5 bài cuối áp dụng tương tự như thế :)