Tính:
a) \(x + 2y + \left( {x - y} \right)\)
b) \(2x - y - \left( {3x - 5y} \right)\)
c) \(3{x^2} - 4{y^2} + 6xy + 7 + \left( { - {x^2} + {y^2} - 8xy + 9x + 1} \right)\)
d) \(4{x^2}y - 2x{y^2} + 8 - \left( {3{x^2}y + 9x{y^2} - 12xy + 6} \right)\)
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1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
a/
\(9x^2+25y^2+1+30xy-6x-10y+4y^2-20y+25=0\)
\(\Leftrightarrow\left(3x+5y-1\right)^2+\left(2y-5\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y-1=0\\2y-5=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{23}{6}\\y=\frac{5}{2}\end{matrix}\right.\)
b/
\(4x^2+4y^2+8xy+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
c/
\(y^2-2y+1+2=\frac{6}{x^2+2x+1+3}\)
\(\Leftrightarrow\left(y-1\right)^2+2=\frac{6}{\left(x+1\right)^2+3}\)
Ta có \(VT=\left(y-1\right)^2+2\ge2\)
\(\left(x+1\right)^2+3\ge3\Rightarrow VP=\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}y-1=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
d/
\(\frac{-9x^2+18x-9-8}{x^2-2x+1+2}=y^2+4y+4-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-8}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-18+10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow-9+\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2+5\)
Ta có \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{10}{\left(x-1\right)^2+2}\le\frac{10}{2}=5\Rightarrow VT\le5\)
\(\left(y+2\right)^2+5\ge5\Rightarrow VP\ge5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2+y^2\right)+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+y^2+\left(x+y-2\right)\left(y+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x^2+y^2-4\right)\left(y+2\right)=-x\left(x^2+y^2\right)\\-\left(x^2+y^2\right)=\left(x+y-2\right)\left(y+2\right)\end{matrix}\right.\)
\(\Rightarrow\left(x^2+y^2-4\right)\left(y+2\right)=x\left(x+y-2\right)\left(y+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\left(\text{không thỏa mãn}\right)\\x^2+y^2-4=x\left(x+y-2\right)\end{matrix}\right.\)
\(\Rightarrow x^2+y^2-4=x^2+x\left(y-2\right)\)
\(\Leftrightarrow\left(y+2\right)\left(y-2\right)=x\left(y-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\x=y+2\end{matrix}\right.\)
Thế vào pt dưới:
\(\Rightarrow\left[{}\begin{matrix}x^2+8+2x+2x-4=0\\\left(y+2\right)^2+2y^2+y\left(y+2\right)+2\left(y+2\right)-4=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Câu b chắc chắn đề sai, nhìn 2 vế pt đầu đều có \(x^2\) thì chúng sẽ rút gọn, không ai cho đề như thế hết
e) Sửa đề: \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=2\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)
PT(1) \(\Leftrightarrow x^3+x\left(x-y^2\right)=\sqrt{\left(x-y^2\right)^3}\)
Đặt \(\sqrt{x-y^2}=a.\text{Thay vào, ta có: }x^3+xa^2-2a^3=0\)
Làm tiếp như ở Câu hỏi của Nguyễn Mai - Toán lớp 9 - Học toán với OnlineMath
Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira, Nguyễn Thị Ngọc Thơ, Nguyễn Lê Phước Thịnh, Quân Tạ Minh, An Võ (leo), @tth_new
e nhiều bài quá giải k kịp mn giúp e vs ạ!cần gấp lắm ạ
thanks nhiều!
a) \(\left(x+3y\right)\left(2x^2y-6xy^2\right)\)
\(=x\left(2x^2y-6xy^2\right)+3y\left(2x^2y-6xy^2\right)\)
\(=2x^3y-6x^2y^2+6x^2y^2-18xy^3\)
\(=2x^3y-18xy^3\)
b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)
\(=6x^5y^2:3x^3y^2-9x^4y^3:3x^3y^2+15x^3y^4:3x^3y^2\)
\(=2x^2-3xy+5y^2\)
c) \(\left(2x+3\right)^2+\left(2x+5\right)^2-2\left(2x+3\right)\left(2x+5\right)\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
d) \(\left(y+3\right)^3-\left(3-y\right)^2-54y\)
\(=y^3+9y^2+27y+27-\left(x^2-6x+9\right)-54y\)
\(=y^3+9y^2-27y+27-x^2+6y-9\)
\(=y^3+9y^2-x^2-21y+18\)
a) \(x+2y+\left(x-y\right)\)
\(=x+2y+x-y\)
\(=2x+y\)
b) \(2x+y-\left(3x-5y\right)\)
\(=2x+y-3x+5y\)
\(=-x+6y\)
c) \(3x^2-4y^2+6xy+7+\left(-x^2+y^2-8xy+9x+1\right)\)
\(=3x^2-4y^2+6xy+7-x^2+y^2-8xy+9x+1\)
\(=2x^2-3y^2-2xy+9x+8\)
d) \(4x^2y-2xy^2+8-\left(3x^2y+9xy^2-12xy+6\right)\)
\(=4x^2y-2xy^2+8-3x^2y-9xy^2+12xy-6\)
\(=x^2y-11xy^2+2+12xy\)