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JS
14 tháng 2 2020
Giải
1) 3xy2 : 5x = \(\frac{3}{5}\)y2
2) 15x4yz3 : 4xyz = \(\frac{15}{4}\)x3z2
3) (4x2y2 - 12xy3 - 7x) : 3x = \(\frac{4}{3}\)xy2 - 4y3 - \(\frac{7}{3}\)
4) (14x4y2 - 12xy3 - x) : 4x = \(\frac{7}{2}\)x3y2 - 3y3 - \(\frac{1}{4}\)
5) (6x2 + 13x - 5) : (2x + 5) = (3x - 1)(2x + 5) : (2x + 5) = 3x - 1
6) (2x4 + x3 - 5x2 - 3x - 3) : (x2 - 3)
= 2x4 + x2 - 6x2 + x3 - 3 - 3x : x2 - 3
= x2(2x2 + x + 1) - 3(2x2 + x + 1) : x2 - 3
= (2x2 + x + 1)(x2 - 3) : x2 - 3
= 2x2 + x + 1
8 tháng 10 2022
a: \(\Leftrightarrow\dfrac{5x^2-13x+6}{A}=\dfrac{5x-3}{2x+5}\)
\(\Leftrightarrow\dfrac{5x^2-10x-3x+6}{A}=\dfrac{5x-3}{2x+5}\)
\(\Leftrightarrow A=\dfrac{\left(2x+5\right)\left(x-2\right)\left(5x-3\right)}{\left(5x-3\right)}=\left(2x+5\right)\left(x-2\right)\)
b: \(\Leftrightarrow\dfrac{x\left(x+4\right)}{A}=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(2x-1\right)}=\dfrac{x}{\left(2x-1\right)}\)
=>A=(x+4)(2x-1)
15 tháng 2 2017
a) Gần giống cho nó giống luôn.
cần thêm (-x^3+2x^2-x) là giống
\(\left(x-1\right)^4+x^3-2x^2+x=\left(x-1\right)^4+x\left(x^2-2x+1\right)=\left(x-1\right)^4+x\left(x-1\right)^2\)
\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=0\\\left(x-1\right)^2+x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
Nghiệm duy nhất: x=1
11 tháng 10 2017
\(\text{a) }x^2+4x+3\\ =x^2+x+3x+3\\ =\left(x^2+x\right)+\left(3x+3\right)\\ =x\left(x+1\right)+3\left(x+1\right)\\ =\left(x+3\right)\left(x+1\right)\\ \)
\(\text{b) }x^2-13x+12\\ =x^2-x-12x+12\\ =\left(x^2-x\right)-\left(12x-12\right)\\ =x\left(x-1\right)-12\left(x-1\right)\\ =\left(x-12\right)\left(x-1\right)\\ \)
\(\text{c) }x^2+5x-6\\ =x^2-x+6x-6\\ =\left(x^2-x\right)+\left(6x-6\right)\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x-1\right)\left(x+1\right)\\ \)
\(\text{d) }2x^2+3x-5\\ =2x^2-2x+5x-5\\ =\left(2x^2-2x\right)+\left(5x-5\right)\\ =2x\left(x-1\right)+5\left(x-1\right)\\ =\left(2x+5\right)\left(x-1\right)\\ \)
\(\text{e) }a^{m+3}-a^m+a-1\\ =\left(a^{m+3}-a^m\right)+\left(a-1\right)\\ =a^m\left(a^3-1\right)+\left(a-1\right)\\ =a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)\\ =\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]\\ =\left(a-1\right)\left(a^{m+2}+a^{m+1}+1\right)\\ \)
11 tháng 10 2017
\(x^2+4x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+3\right)\left(x+1\right)\)
\(x^2-13x+12=x\left(x-12\right)-\left(x-12\right)=\left(x-12\right)\left(x-1\right)\)
\(x^2+3x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x-2\right)\left(x+5\right)\)
\(x^2+5x-6=x\left(x-6\right)+\left(x-6\right)=\left(x-6\right)\left(x+1\right)\)
\(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
\(a^{m+3}-a^m+a-1=a^m\left(a^3-1\right)+\left(a-1\right)=a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)=\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]=\left(a-1\right)\left(a^{m+2}+a^{m+1}+a^m+1\right)\)
6 tháng 7 2018
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
25 tháng 11 2016
a)\(6x^2+5x-6=0\)
\(\Leftrightarrow6x^2-4x+9x-6=0\)
\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
b)\(6x^2-13x+6=0\)
\(\Leftrightarrow6x^2-4x-9x+6=0\)
\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
c)\(10x^2-13x-3=0\)
\(\Leftrightarrow10x^2-15x+2x-3=0\)
\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)
d)\(20x^2+19x-3=0\)
\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)
\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)
e)\(3x^2-x+6=0\)
\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)
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