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2 tháng 7 2021
1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
Vậy.....
2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy...
3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
CT
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2023
ĐKXĐ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
29 tháng 10 2023
\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\)
\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\cdot1\)
\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
NN
2
KT
12 tháng 2 2018
a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow\)\(2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(5x+4\right)=0\)
đến đây tự lm nha
b) \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\) (1)
ĐKXĐ: \(x\ne\pm\frac{1}{3}\)
\(\left(1\right)\)\(\Leftrightarrow\)\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
\(\Rightarrow\)\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)
\(\Leftrightarrow\)\(\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)
\(\Leftrightarrow\)\(-12x=12\)
\(\Leftrightarrow\)\(x=-1\) (t/m ĐKXĐ)
Vậy....
12 tháng 2 2018
a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{4}{5}\end{cases}}}\)
b) ĐKXĐ: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
\(\Leftrightarrow\left(1-3x\right)^2-\left(1+3x\right)^2=12\)
\(\Leftrightarrow\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)
\(\Leftrightarrow-12x=12\)
\(\Leftrightarrow x=-1\) (thỏa mãn)
Vậy x = -1
4 tháng 3 2022
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
4 tháng 3 2022
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
11 tháng 4 2020
a/2(9x2+6x+1)=(3x+1)(x-2)
⇔2(3x+1)2= (3x+1)(x-2)
⇔ 2(3x+1)2 :(3x+1)=x-2
⇔ 2(3x+1)=x-2
⇔6x+2-x+2=0
⇔5x+4=0
⇔5x=-4
⇔x=\(\frac{-4}{5}\)
b/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
⇔\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
⇔12=(1-3x)2-(1+3x)2
⇔-(1-3x-1-3x)(1-3x+1+3x)=--12
⇔-(-6x.2)=-12
⇔12x=-12
⇔x=-1
bạn thấy mình làm sai hay thiếu thì bạn nhớ nhắc mình nha.
BN
0
TA
17 tháng 6 2023
\(3x\left(x-1\right)+\left(x+1\right)^2=1+3x^2\)
\(\Leftrightarrow3x^2-3x+x^2+2x+1=1+3x^2\)
\(\Leftrightarrow3x^2+x^2-3x^2-3x+2x+1-1=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{0;1\right\}\).
15 tháng 3 2022
\(\left|2x+1\right|=4.\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=-4.\\2x+1=4.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}.\\x=\dfrac{3}{2}.\end{matrix}\right.\)
\(\left|3x-2\right|+1=0.\)
\(\Leftrightarrow\left|3x-2\right|=-1\) (vô lý).
\(\Rightarrow x\in\phi.\)
Nếu \(x\ge\frac{1}{3}\) thì \(\left|3x-1\right|=3x-1\)
Khi đó pt trở thành \(1-3x+1=2\)
\(\Leftrightarrow\)\(-3x=0\)
\(\Leftrightarrow\) \(x=0\) (loại)
Nếu \(x< \frac{1}{3}\)thì \(\left|3x-1\right|=1-3x\)
Khi đó phương trình trở thành: \(1-1+3x=2\)
\(\Leftrightarrow\)\(3x=2\)
\(\Leftrightarrow\) \(x=\frac{2}{3}\) (loại)
Vậy....
\(|3x-1|>=0\)với mọi x\(\Rightarrow1-|3x-1|< =1-0=1\Rightarrow1-|3x-1|< =1\)
\(\Rightarrow1-|3x-1|=2\)vô lí \(\Rightarrow x\in\varnothing\)
vậy .....