Tìm x:
a) 5(2y+3)(y+2) - 2(5y-4)(y-1) = 75
b) (2x+3)^2 + (2-2x)^2 + (4x+6)(2-2x) = x+1
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a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
\(\hept{\begin{cases}3x=2y\\2x+y=3\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{3}{2}.x\\2x+\frac{3}{2}.x=3\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{3}{2}.x\\\frac{7}{2}.x=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{6}{7}\\y=\frac{9}{7}\end{cases}}}\)
\(\hept{\begin{cases}\frac{x}{3}=\frac{3y}{4}\\3x-y=4\end{cases}\Leftrightarrow\hept{\begin{cases}4x=9y\\3x-y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9y}{4}\\\frac{3.9}{4}y-y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{4}.y\\\frac{23}{4}.y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{4}.y\\y=\frac{16}{23}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{36}{23}\\y=\frac{16}{23}\end{cases}}}\)
Các phần sau làm tương tự nhé
Câu a tự làm nhé
b, \(\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Leftrightarrow32(2x+3)=24(3x-1)\)
\(\Leftrightarrow64x+96=72x-24\)
\(\Leftrightarrow64x+96-72x=-24\)
\(\Leftrightarrow96-8x=-24\Leftrightarrow x=15\)
1) \(\left(x+1\right)^2-16y^2\)
\(=\left(x+1\right)^2-\left(4y\right)^2\)
\(=\left(x+1-4y\right).\left(x+1+4y\right)\)
2) \(2x^2-8\)
\(=\left[\left(\sqrt{2x}\right)^2\right]^2-\left(\sqrt{8}\right)^2\)
\(=\left[\left(\sqrt{2x}\right)^2-\sqrt{8}\right].\left[\left(\sqrt{2x}\right)^2+\sqrt{8}\right]\)
6) \(x^2+1-y^2+2x\)
\(=x^2+2x+1-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right).\left(x+1+y\right)\)
Chúc bạn học tốt!
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]
a) 5( 2y + 3 )( y + 2 ) - 2( 5y - 4 )( y - 1 ) = 75
<=> 5( 2y2 + 7y + 6 ) - 2( 5y2 - 9y + 4 ) = 75
<=> 10y2 + 35y + 30 - 10y2 + 18y - 8 = 75
<=> 53y + 22 = 75
<=> 53y = 53
<=> y = 1
b) ( 2x + 3 )2 + ( 2 - 2x )2 + ( 4x + 6 )( 2 - 2x ) = x + 1
<=> 4x2 + 12x + 9 + 4x2 - 8x + 4 - 8x2 - 4x + 12 = x + 1
<=> 25 = x + 1
<=> x = 24