(x+1)(3x+4)(6x+7)2 = 6
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KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
23 tháng 7 2018
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
31 tháng 7 2022
a: \(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
=>14x=7
hay x=1/2
b: \(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
=>-56x+156=24x-324
=>-80x=-480
hay x=6
c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2
HN
22 tháng 8 2021
\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)
\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=n\)
\(=\left(n+3\right)n-4\)
\(=n^2+3n-4\)
\(=n^2-n+4n-4\)
\(=n\left(n-1\right)+4\left(n-1\right)\)
\(=\left(n-1\right)\left(n+4\right)\)
\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)
\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
HN
22 tháng 8 2021
\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)
\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)
Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):
\(\left(12n+1\right)n=6\)
\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)
26 tháng 10 2023
6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
NN
4
K
16 tháng 4 2022
\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{3}{7}\)
\(x=\dfrac{7}{9}\)
\(x:\dfrac{6}{8}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}\times\dfrac{6}{8}\)
\(x=\dfrac{6}{20}=\dfrac{3}{10}\)
\(x-\dfrac{2}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{2}{3}\)
\(x=\dfrac{9}{6}=\dfrac{3}{2}\)
\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)
\(x=\dfrac{1}{7}+\dfrac{4}{35}+\dfrac{2}{5}\)
\(x=\dfrac{23}{35}\)
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LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
\(\left(x+1\right)\left(3x+4\right)\left(6x+7\right)^2=6\)
<=> \(6\left(x+1\right).2\left(3x+4\right)\left(6x+7\right)^2=72\)
<=> \(\left(6x+6\right)\left(6x+8\right)\left(6x+7\right)^2-72=0\) (*)
Đặt: \(6x+7=t\) khi đó pt (*) trở thành:
\(\left(t-1\right)\left(t+1\right)t^2-72=0\)
<=> \(t^4-t^2-72=0\)
<=> \(\left(t-3\right)\left(t+3\right)\left(t^2+3\right)=0\)
<=> \(\orbr{\begin{cases}t-3=0\\t+3=0\end{cases}}\) (do t2 + 3 > 0 )
<=> \(\orbr{\begin{cases}t=3\\t=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}6x+7=3\\6x+7=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{cases}}}\)
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