1, thực hiện phép tính a, ( x+10).(x-10) b, (x-3)(x²+3x+9)
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CM
29 tháng 11 2019
a) Rút gọn thu được kết quả: 3;
b) Ta có MC = 3x (x - 3)
Thực hiện tính toán thu được kết quả: x 2 − 6 x + 9 3 x ( x − 3 ) = x − 3 3 x
c) Trước tiên biến đổi: 3 + 3 x = 3 ( x + 1 ) x ; 3 3 ( x + 1 ) x = x x + 1
Thay vào A và thu gọn ta được A = 4 x + 3 x
2 tháng 12 2023
Bài 4:
1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)
=>\(x^3-1-x^3-6x=11\)
=>-6x-1=11
=>-6x=11+1=12
=>\(x=\dfrac{12}{-6}=-2\)
2: \(16x^2-\left(3x-4\right)^2=0\)
=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)
=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)
=>(x+4)(7x-4)=0
=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)
3: \(x^3-x^2-3x+3=0\)
=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)
=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-3\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))
=>\(\left(x+2\right)^2=\left(x-1\right)\left(x+1\right)\)
=>\(x^2+4x+4=x^2-1\)
=>4x+4=-1
=>4x=-5
=>\(x=-\dfrac{5}{4}\left(nhận\right)\)
5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)
=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)
=>3x+1=0
=>3x=-1
=>\(x=-\dfrac{1}{3}\left(nhận\right)\)
6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)
\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)
=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-x-3}{x}=1\)
=>-x-3=x
=>-2x=3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
10 tháng 1 2023
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
13 tháng 12 2018
\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)
\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{1}{2\left(x-3\right)}\)
\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)
\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)
\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)
\(=\frac{12}{x}\)
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
a) \(\dfrac{4n^2}{17n^4}\cdot\dfrac{-7n^2}{12n}\) \(\left(n\ne0\right)\)
\(=\dfrac{4n^2\cdot-7n^2}{17n^4\cdot12n}\)
\(=\dfrac{-28n^4}{204n^5}\)
\(=\dfrac{-7}{51n}\)
b) \(\dfrac{3x-1}{10x^2+2x}\cdot\dfrac{25x^2+10x+1}{1-9x^2}\) \(\left(x\ne\pm\dfrac{1}{3};x\ne0;x\ne-\dfrac{1}{5}\right)\)
\(=\dfrac{3x-1}{2x\left(5x+1\right)}\cdot\dfrac{\left(5x+1\right)^2}{\left(1-3x\right)\left(3x+1\right)}\)
\(=\dfrac{-\left(1-3x\right)\left(5x+1\right)^2}{2x\left(5x+1\right)\left(1-3x\right)\left(1+3x\right)}\)
\(=\dfrac{-\left(5x+1\right)}{2x\left(1+3x\right)}\)
\(=-\dfrac{5x+1}{6x^2+2x}\)
c) \(\dfrac{27-a^3}{5a+10}:\dfrac{a-3}{3a+6}\) \(\left(a\ne-2;a\ne3\right)\)
\(=\dfrac{\left(3-a\right)\left(9+3a+a^2\right)}{5\left(a+2\right)}\cdot\dfrac{3\left(a+2\right)}{a-3}\)
\(=\dfrac{-\left(a-3\right)\left(a^2+3a+9\right)\cdot3\left(a+2\right)}{5\left(a+2\right)\left(a-3\right)}\)
\(=\dfrac{-3\left(a^2+3x+9\right)}{5}\)
\(=-\dfrac{3x^2+9x+27}{5}\)
d) \(\dfrac{x^2-1}{x^2+2x-15}:\dfrac{x^2+5x+4}{x^2-10x+21}\) \(\left(x\ne3;x\ne-5;x\ne-1;x\ne-4\right)\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}:\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-3\right)\left(x-7\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}\cdot\dfrac{\left(x-3\right)\left(x-7\right)}{\left(x+1\right)\left(x+4\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)
PH
0
a: =x^2-100
b: =x^3-27