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NN
6 tháng 12 2017
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-6\)
LH
6 tháng 12 2017
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=6\)
27 tháng 6 2019
\(A=-2x^2+5x-8\)
\(A=-2\left(x^2-\frac{5}{2}\cdot x+4\right)\)
\(A=-2\left(x^2-2\cdot x\cdot\frac{5}{4}+\frac{25}{16}+\frac{39}{16}\right)\)
\(A=-2\left[\left(x-\frac{5}{4}\right)^2+\frac{39}{16}\right]\)
\(A=-2\left(x-\frac{5}{4}\right)^2-\frac{39}{6}\le\frac{-39}{6}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{5}{4}\)
\(B=-x^2-y^2+xy+2x+2y\)
\(2B=-2x^2-2y^2+2xy-4x-4y\)
\(2B=-\left(2x^2+2y^2-2xy+4x+4y\right)\)
\(2B=-\left(x^2-2xy+y^2+x^2+4x+4+y^2+4y+4-8\right)\)
\(2B=-\left[\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2-8\right]\)
\(B=-\frac{\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2}{2}+4\le4\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=-2\)
\(C=\frac{3}{4x^2-4x+5}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
\(D=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}\)
\(=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}=\frac{5}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
PN
13 tháng 6 2019
a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)-3=-3\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3-3=-3\)
\(\Leftrightarrow14x=0\)
\(\Leftrightarrow x=0\)
Vậy pt có nghiệm duy nhất x = 0.
b) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=7\)
\(\Leftrightarrow18x-2=7\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy pt có nghiệm duy nhất \(x=\frac{1}{2}\)
c) \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow36x^2-24x+4+25x^2-20x+4-60x^2+33x-8=0\)
\(\Leftrightarrow x^2-11x=0\)
\(\Leftrightarrow x\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;11\right\}\)
d) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-4x+32=1\)
\(\Leftrightarrow41-10x=1\)
\(\Leftrightarrow-10x=40\)
\(\Leftrightarrow x=-4\)
Vậy pt có nghiệm duy nhất x = -4.
e) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7x^2+36=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2=0\)
\(\Leftrightarrow8x=-13\)
\(\Leftrightarrow x=-\frac{13}{8}\)
Vậy pt có nghiệm duy nhất \(x=-\frac{13}{8}\)
30 tháng 11 2017
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
Những câu khác làm tương tự
\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
=> \(\left(2x+3-2x+5\right)^2=x^2+6x+64\)
=> \(8^2=x^2+6x+64\)
=> \(64=x^2+6x+64\)
=> \(x\left(x+6\right)=0\)
=> \(\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)
\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2=x^2+6x+64\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)
\(\Leftrightarrow8^2=x^2+6x+64\)\(\Leftrightarrow64=x^2+6x+64\)
\(\Leftrightarrow x^2+6x+64-64=0\)\(\Leftrightarrow x^2+6x=0\)
\(\Leftrightarrow x\left(x+6\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{-6;0\right\}\)