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1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
Bài 1:
a: \(6x^2-11x+3\)
\(=6x^2-9x-2x+3\)
\(=3x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(3x-1\right)\)
b: \(2x^2+3x-27\)
\(=2x^2+9x-6x-27\)
\(=x\left(2x+9\right)-3\left(2x+9\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)
c: \(x^2-10x+24\)
\(=x^2-4x-6x+24\)
\(=x\left(x-4\right)-6\left(x-4\right)\)
\(=\left(x-4\right)\left(x-6\right)\)
d: \(49x^2+28x-5\)
\(=49x^2+28x+4-9\)
\(=\left(7x+2\right)^2-9\)
\(=\left(7x-1\right)\left(7x+5\right)\)
e: \(2x^2-5xy-3y^2\)
\(=2x^2-6xy+xy-3y^2\)
\(=2x\left(x-3y\right)+y\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x+y\right)\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
\(a,\left(2x-1\right)^2=49\)
\(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(b,\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(4x^2+28x+49=9x^2+36x+36\)
\(4x^2+28x+49-9x^2-36x-36=0\)
\(-5x^2-8x+13=0\)
\(5x^2+13-5x-13=0\)
\(x\left(5x+13\right)-1\left(5x+13\right)=0\)
\(\left(x-1\right)\left(5x+13\right)=0\)
\(\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=-\frac{13}{5}\end{matrix}\right.\)
\(c,4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(x=-5\)
\(d,\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
\(25x^2-30x+9-16x^2+56x-49=0\)
\(9x^2+26x-40=0\)
\(9x^2+36x-10x-40=0\)
\(9x\left(x+4\right)-10\left(x+4\right)=0\)
\(\left(9x-10\right)\left(x+4\right)=0\)
\(\left[{}\begin{matrix}9x-10=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{10}{9}\\x=-4\end{matrix}\right.\)
B1:
a) \(9x^2+90x+225-\left(x-7\right)^2\)
= \(9x^2+90x+225-x^2+14x-49\)
= \(8x^2+104x+176\)
= \(\left(x+2\right)\left(x+11\right)\)
b) \(49\left(y-4\right)^2-9y^2-36y+36\)
= \(49\left(y^2-8y+16\right)-9y^2-36y+36\)
= \(49y^2-392y+784-9y^2-36y+36\)
= \(40y^2-428y+820\)
= \(\left(5y-41\right)\left(8y-20\right)\)
B2:
a) A = \(xy-4y-5y+20=xy-9y+20\)
A = \(y\left(x-9\right)+20\)
Với x = 14, y = \(\dfrac{11}{2}\)
A = \(\dfrac{11}{2}\left(14-9\right)+20=47,5\)
b) B = \(x^2+xy-5x-5y\)
B = \(x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Với x = -5, y = -8
B = \(\left(-5-8\right)\left(-5-5\right)=130\)
B3:
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\left(2x-5\right)\left(-2\right)=0\)
\(x=\dfrac{5}{2}\)
b) \(\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\left(x+3\right)x\left(x-2\right)=0\)
\(\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
c) \(\left(2x^3+2x^2\right)+\left(3x^2+3\right)=0\)
\(2x^3+5x^2+3=0\)
\(\Rightarrow\) Đề sai rồi, nghiệm khủng bố lắm.
a)\(\left(2x-y\right)[\left(2x\right)^2-2.2x.y+y^2]\)
\(=\left(2x-y\right)^3\)
b)\(2x^2-3xy+5y^2\)
c)\(2x^3-10x^2-11x^2+55x+12x-60\)
\(=2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\)
\(=\left(x+5\right)\left(2x^2-11x+12\right)\)
\(\Leftrightarrow(2x^3-21x^2+67x-60)/\left(x-5\right)=2x^2-11x+12\)
mơn bạn nha