giúp mik với mn ơi mik đg cần gấp
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giúp mik với mik đg cần gấp
14 tháng 11 2023
a: \(\left\{{}\begin{matrix}3x-2y=5\\-2x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-2y=5\\-4x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x=11\\-2x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-11\\y=2x+3=-22+3=-19\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}2x-y=4\\x-\dfrac{y}{2}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y=4\\2x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=0\left(luônđúng\right)\\2x-y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\in R\\2x=y+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\in R\\x=\dfrac{1}{2}y+2\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}y\in R\\x=\dfrac{y+4}{2}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}3x+2y=-2\\5x+4y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+4y=-4\\5x+4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-5x=-4-1=-5\\5x+4y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\4y=1-5x=1+25=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{26}{4}=\dfrac{13}{2}\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}2x-y=6\\3x+5y=22\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}10x-5y=30\\3x+5y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=52\\2x-y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\2x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2x-6=2\cdot4-6=2\end{matrix}\right.\)
e: \(\left\{{}\begin{matrix}-x+2y-6=0\\5x-3y-5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+10y=30\\5x-3y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7y=35\\x-2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=2y-6=10-6=4\end{matrix}\right.\)
g: \(\left\{{}\begin{matrix}2x-3y=8\\5x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-6y=16\\15x+6y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=19\\2x-3y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\3y=2x-8=2-8=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
6 tháng 8 2023
a: \(Q=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{x-4}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)
\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\cdot\left(-1\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b: Khi x=4-2căn 3 thì \(Q=\dfrac{\sqrt{3}-1+2}{\sqrt{3}-1-3}=\dfrac{\sqrt{3}+1}{\sqrt{3}-4}=\dfrac{-7-5\sqrt{3}}{13}\)
c: Q>1/6
=>Q-1/6>0
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{1}{6}>0\)
=>\(\dfrac{6\sqrt{x}+12-\sqrt{x}+3}{6\left(\sqrt{x}-3\right)}>0\)
=>\(\dfrac{5\sqrt{x}+9}{6\left(\sqrt{x}-3\right)}>0\)
=>căn x-3>0
=>x>9
giúp mik với đg cần gấp
giúp mik với đg cần gấp
giúp mik với đg cần gấp
giúp mik với đg cần gấp
6 tháng 11 2023
a: Thay x=2 và y=0 vào (d), ta được:
2(2-m)+m+1=0
=>4-2m+m+1=0
=>5-m=0
=>m=5
b: Đề sai rồi bạn
c: Thay x=2 vào y=-2x+3, ta được:
\(y=-2\cdot2+3=-4+3=-1\)
Thay x=2 và y=-1 vào (d), ta được:
2(2-m)+m+1=-1
=>4-2m+m+1=-1
=>-m+5=-1
=>-m=-6
=>m=6
d: Thay y=-2 vào y=2x-3, ta được:
2x-3=-2
=>2x=1
=>x=1/2
Thay x=1/2 và y=-2 vào (d), ta được:
\(-\dfrac{1}{2}\left(2-m\right)+m+1=-2\)
=>\(-1+\dfrac{1}{2}m+m+1=-2\)
=>\(\dfrac{3}{2}m=-2\)
=>\(m=-2:\dfrac{3}{2}=-\dfrac{4}{3}\)
e: Thay x=1 và y=2 vào (d), ta được:
\(1\left(2-m\right)+m+1=2\)
=>2-m+m+1=2
=>3=2(vô lý)
5B
Mn giúp mik với,bh mik đang cần gấp lắm 19h30 mik phải có bài r nên mn giúp mik với
6
7:
a: ĐKXĐ: x>=0; x<>1
\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)
b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)
c: |D|=1/3
=>D=-1/3 hoặc D=1/3
=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)
=>\(\sqrt{x}+1=3\)
=>\(\sqrt{x}=2\)
=>x=4
6:
a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
b: C<-1
=>C+1<0
=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)
=>\(-\sqrt{x}+4< 0\)
=>\(-\sqrt{x}< -4\)
=>\(\sqrt{x}>4\)
=>x>16
\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để `C < -1` Ta có :
\(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)