3x.3x+1.3x+2=36
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Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
\(3x^2-3x=36\)
\(\Leftrightarrow x^2-x=12\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(3x-12\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
Dễ rồi tự làm
\(3x^2-3x\left(-2+x\right)=36\)
\(=3x^2+6x-3x=36\)
\(=3x^2-3x=36\)
\(=3x\left(x-3\right)=36\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=36\\x-3=36\end{cases}\Leftrightarrow\orbr{\begin{cases}x=36:3=12\\x=36+3=39\end{cases}}}\)
Vậy S = { 12 ; 39 }
\(3x^{2}+3x-36\)
\(=x^{2}+x-12\)
\(=x^{2}-3x+4x-12\)
\(=x(x-3)+4(x-3)\)
\(=(x+4)(x-3)\)
\(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
A=\(\dfrac{3x^3-14x^2+3x+36}{3x^3-19x^2+33x-9}\)
=>A \(=\dfrac{\left(x-3\right)\left(3x^2-5x-12\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
=>A=\(\dfrac{\left(x-3\right)^2\left(3x+4\right)}{\left(x-3\right)^2\left(3x-1\right)}\)
=>A=\(\dfrac{3x+4}{3x-1}\)
a) \(3x+2\left(5-x\right)=-11\)
\(\Leftrightarrow3x+10-2x=-11\)
\(\Leftrightarrow x=-21\)
b) \(3x^2-3x\left(x-2\right)=36\)
\(\Leftrightarrow3x^2-3x^2+6x=36\)
\(\Rightarrow x=6\)
3x+x+1+x+2=33x+3=32.22
=>x ko có giá trị
\(3^x.3^{x+1}.3^{x+2}=36\)
\(\Rightarrow\)\(3^{x+x+1+x+2}=36\)
\(\Rightarrow\)\(3^{3x+3}=36\)
\(\Rightarrow\) \(3^{3x}\cdot3^3=36\)
\(\Rightarrow\) \(3^{3x}.27=36\)
\(\Rightarrow\) \(3^{3x}=\frac{4}{3}\)
\(\Rightarrow\)Không có giá trị x thỏa mãn