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\(2x^2+y^2+2xy-8x-6y+10=0\)
\(\Rightarrow2.\left(2x^2+y^2+2xy-8x-6y+10\right)=0\)
\(\Rightarrow4x^2+2y^2+4xy-16x-12y+20=0\)
\(\Rightarrow\left(4x^2+y^2+16+4xy-8y-16x\right)+\left(y^2-4y+4\right)=0\)
\(\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2=0\left(1\right)\)
Ta có: \(\hept{\begin{cases}\left(2x+y-4\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2\ge0\forall x;y\left(2\right)}\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+y=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}2x+2=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
Chúc bạn học tốt.
\(A=x^2+10y^2+2x-6xy-10y+25\)
=> \(A=x^2+2x\left(1-3y\right)+\left(1-3y\right)^2-\left(1-3y\right)^2-10y+25\)
=> \(A=\left(x+1-3y\right)^2-1+6y-9y^2-10y+25\)
=> \(A=\left(x+1-3y\right)^2-9y^2-4y+24\)
=> \(A=\left(x+1-3y\right)^2-\left(3y\right)^2-2.3y.\frac{2}{3}-\left(\frac{2}{3}\right)^2+\frac{220}{9}\)
=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\)
Có \(\left(x+1-3y\right)^2\ge0\)với mọi x, y
\(\left(3y+\frac{2}{3}\right)^2\ge0\)với mọi y
=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\ge\frac{220}{9}\)với mọi x, y
Dấu "=" xảy ra <=> \(\left(x+1-3y\right)^2=0\)<=> \(x+1-3y=0\)
và \(\left(3y+\frac{2}{3}\right)^2=0\)=> \(3y+\frac{2}{3}=0\)
=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)
Bổ xung phần kết luận
KL: Amin = \(\frac{220}{9}\)<=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)
1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)
Vậy ...................
b/ ĐKXĐ:\(x\ne2;x\ne5\)
.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)
\(\Leftrightarrow2x^2-10x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)
Vậy ..............
`Answer:`
`1.`
a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)
b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)
\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)
\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)
`2.`
\(ĐKXĐ:x\ne-m-2;x\ne m-2\)
Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)
a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)
b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì
\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)
Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)
\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2}\)
2. Tìm x:
( x - 3 )2 - x + 3 = 0
=> x2 - 6x + 9 - x + 3 = 0
=> x2 - 7x + 12 = 0
=> ( x2 - 3x ) + ( 4x - 12 ) = 0
=> x.(x - 3) + 4.(x - 3) = 0
=> ( x - 3 ).( x + 4 ) = 0
=> x - 3 = 0 => x = 3
x + 4 = 0 => x = -4
Trl:
1.
a. \(75^2+150\text{.}25+25^2\)
\(=75^2+2\text{.}75\text{.}25+25^2\)
\(=\left(75+25\right)^2\)
\(=100^2\)
\(=10000\)
b. \(2019^2-2019.19-19^2-19.1981\)
(Đề bài có sai ko vậy???)~ hoặc lak do mk ngu quá k bt lm
2. \(\left(\text{x}-3\right)^2-\text{x}+3=0\)
\(\text{x}^2-6\text{x}+9-\text{x}+3=0\)
\(\text{x}^2-7\text{x}+12=0\)
\(\text{x}^2-3\text{x}-4\text{x}+12=0\)
\(\text{x}\left(\text{x}-3\right)-4\left(\text{x}-3\right)=0\)
\(\left(\text{x}-3\right)\left(\text{x}-4\right)=0\)
\(\orbr{\begin{cases}\text{x}-3=0\\\text{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\text{x}=3\\\text{x}=4\end{cases}}}\)
Vậy ....
#HuyềnAnh#