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23 tháng 2 2020
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
MT
0
8 tháng 1 2020
a) 9x2 - 1 = (3x + 1)(2x - 3)
=> 9x2 - 1 = 6x2 - 9x + 2x - 3
=> 9x2 - 6x2 + 7x - 1 + 3 = 0
=> 3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=>\(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
b) 2(9x2 + 6x + 1) = (3x + 1)(x - 2)
=> 2(3x + 1)2 - (3x + 1)(x - 2) = 0
=> (3x + 1)(6x + 2 - x + 2) = 0
=> (3x + 1)(5x +4 ) = 0
=> \(\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-\frac{4}{5}\end{cases}}\)
c) 27x2(x + 3) - 12(x2 + 3x) = 0
=> 27x2(x + 3) - 12x(x + 3) = 0
=> 3x(9x - 4)(x + 3) = 0
=> 3x = 0
9x - 4 = 0
x + 3 = 0
=> x = 0
x = 4/9
x = -3
d) 16x2 - 8x + 1 = 4(x + 3)(4x - 1)
=> (4x - 1)2 - 4(x + 3)(4x - 1) = 0
=> (4x - 1)(4x - 1 - 4x - 12) = 0
=> 4x - 1 = 0
=> x = 1/4
6 tháng 8 2016
a) \(\left(3x-4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)
\(\Leftrightarrow27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)
\(\Leftrightarrow27x^3+108x^2+144x+64-27x^3+72x+24x^2-64=0\)
\(\Leftrightarrow132x^2+216x=0\)
\(\Leftrightarrow12x\left(11x+18\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\11x+18=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{18}{11}\end{array}\right.\)
28 tháng 7 2019
b) ta có \(\left(4x-5\right)^3-\left(2x+5\right)\left(16x^2-25\right)=0\)
\(\left(4x-5\right)^3-\left(2x+5\right)\left(4x+5\right)\left(4x-5\right)=0\)
\(\left(4x-5\right)\left[\left(4x-5\right)^2-\left(2x+5\right)\left(4x+5\right)\right]=0\)
\(\left(4x-5\right)\left(16x^2-40x+5^2-8x^2-10x-20x-5^2\right)=0\)
\(\left(4x-5\right)\left(8x^2-70x\right)=0\)
=> \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}4x=5\\x\left(8x-70\right)=0\end{cases}< =>}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}\) \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}}\)
28 tháng 7 2019
\(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{35}{4}\end{cases}}\) Vậy \(\orbr{\begin{cases}x=\frac{5}{4}\\x=0\end{cases}}\) hoặc \(x=\frac{35}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 10 2019
\(A=\left(5x\right)^2-1^2=\left(5x-1\right)\left(5x+1\right)\)
\(B=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
\(C=x\left(x^2+8x+16\right)=x\left(x+4\right)^2\)
\(D=\left(x-y\right)\left(x+y\right)-3\left(x+y\right)=\left(x-y-3\right)\left(x+y\right)\)
\(E=\left(3x\right)^2-\left(y^2-10y+25\right)=\left(3x\right)^2-\left(y-5\right)^2\)
\(=\left(3x-y+5\right)\left(3x+y-5\right)\)
\(F=x^2+x+7x+7=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
\(G=3x^2+3x-5x-5=3x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(3x-5\right)\)
\(H=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)