a. 2 (x-1) + 6x = 12- 3x
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18 tháng 12 2022
a: =18x^3y^2-12x^3y^3+6x^2y^2
b: (-3x+2)(5x^2-1/3x+4)
=-12x^3+x^2-12x+10x^2-2/3x+8
=-12x^3+11x^2-38/3x+8
c: =x^2-x-2+3x-x^2
=2x-2
d: =4x^2+12x+9-4x^2+25-(x-1)(x^2+12)
=12x+34-x^3-12x+x^2+12
=-x^3+x^2+46
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
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.
CN
1
22 tháng 11 2018
a) 6x(2x-4)+4(9-3x2)=-12
12x2-24x+36-12x2=-12
-24x+36=-12
-24x=-12-36
-24x=-48
x=2
b) 9x2-(3x+1)(4x-5)+1+6x=0
9x2-(3x+1)(4x-5)+1+6x=0
9x2-(12x2-11x-5)+1+6x=0
9x2-12x2+11x+5+1+6x=0
-3x2-17x-6=0
\(x=\dfrac{-\left(-17\right)+-\sqrt{\left(-17\right)^2-4.3\text{x}\left(-6\text{x}\right)}}{2.3}\)
\(x=\dfrac{17+-\sqrt{289+72}}{6}\)
\(x=\dfrac{17+-\sqrt{361}}{6}\)
\(x=\dfrac{17+-19}{6}\)
\(x=\dfrac{17+19}{6}\)
\(x=\dfrac{17-19}{6}\)
x=6
\(x=\dfrac{-1}{3}\)
17 tháng 3 2023
a: \(T=\dfrac{3}{2}x^4-x^3+3x^2-\dfrac{1}{2}x+6+x^4+\dfrac{2}{3}x^3-2x^2-4x+1\)
\(=\dfrac{5}{2}x^4-\dfrac{1}{3}x^3+x^2-\dfrac{9}{2}x+7\)
b: \(T\left(2\right)=\dfrac{5}{2}\cdot16-\dfrac{1}{3}\cdot8+4-\dfrac{9}{2}\cdot2+7=\dfrac{118}{3}\)
\(2\times\left(x-1\right)+6x=12-3x\)
\(2x-2+6x=12-3x\)
\(2x+6x+3x=12+2\)
\(x\times\left(2+6+3\right)=14\)
\(11x=14\)
\(x=\frac{14}{11}\)
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dạ là câu B ạ
B.Hướng "phi tập trung"