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1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
=>x=0
2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5=7\)
=>6x-12=7
=>6x=19
hay x=19/6
1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)-3=-6\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=-3\)
\(\Leftrightarrow14x=-3\)
hay x=-3/14
2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5\)
=>4x-12=7
=>4x=19
hay x=19/4
Câu a : \(\left(2x-3\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x+2\right)\left(2x-3+x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy ........
b ) \(\left(3x+1\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=0\)
\(\Leftrightarrow5x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy.............
c ) \(x^3+2x^2+6x+12=0\)
\(\Leftrightarrow x^2\left(x+2\right)+6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x\left(loại\right)\end{matrix}\right.\) Do \(x^2+6>0\)
Vậy.........
a)\(\left(2x-3\right)^2=\left(x-2\right)^2\)
\(2x-3=x-2\)
\(2x-3-x+2=0\)
\(x-1=0\)
\(x=1\)
b)\(\left(3x+1\right)^2=\left(2x-1\right)^2\)
\(3x+1=2x-1\)
\(3x+1-2x+1=0\)
\(x+2=0\)
\(x=-2\)
c)\(x^3+2x^2+6x+12=0\)
\(\left(x^3+2x^2\right)+\left(6x+12\right)=0\)
\(x^2\left(x+2\right)+6\left(x+2\right)=0\)
\(\left(x^2+6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-6\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{6}\\x=-2\end{matrix}\right.\)
Vậy \(x=-\sqrt{6}\) hoặc \(x=-2\)
1. <=> \(\left(3x+2\right)^3-\left(\left(3x\right)^3+2^3\right)=0\)
<=> \(\left(\left(3x\right)^3+2^3+3\left(3x+2\right).3x.2\right)-\left(\left(3x\right)^3+2^3\right)=0\)
<=>3 (3x + 2) . 3x.2 = 0
<=> (3x + 2 ) . x = 0
<=> x = -2/3 hoặc x = 0
2. Tương tự
1
\(\left(3x+2\right)^3-\left[\left(3x\right)^3+2^3\right]=0\)
\(\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2+2^3-\left(3x\right)^3-2^3=0\)
\(54x^2+36x=0\)
\(18x\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{-2}{3}\end{cases}}\)
2
\(\left(2x+1\right)^3-\left[\left(2x\right)^3-1^3\right]=0\)
\(\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3-\left(2x\right)^3-1^3=0\)
\(12x^2+6x=0\)
\(6x\left(2x+1\right)=0\)
\(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
\(x^4-3x^3-6x^2+3x+1\)
\(=x^4+x^3-x^2-4x^3-4x^2+4x-x^2-x+1\)
\(=\left(x^2+x-1\right)\left(x^2-4x-1\right)\)