Gọi thời gian làm một mình của tổ 1 là: `x` (giờ)

Thời gian làm một mình của tổ 2 là: `y` (giờ)

ĐK: `x,y>0` 

Hai tổ cùng làm thì 15h xong nên ta có pt: `1/x+1/y=1/15` (1) 

Nếu tổ 1 làm trong 3h và tổ 2 làm trong 5h thì được 25% công việc nên ta có pt: 

\(\dfrac{3}{x}+\dfrac{5}{y}=\dfrac{1}{4}\left(2\right)\)

Từ (1) và (2) ta có hpt: 

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{15}\\\dfrac{3}{x}+\dfrac{5}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{15}\\3\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{2}{y}=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{15}\\\dfrac{1}{5}+\dfrac{2}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{15}\\\dfrac{2}{y}=\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{1}{20}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{40}=\dfrac{1}{15}\\y=2:\dfrac{1}{20}=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{15}-\dfrac{1}{40}=\dfrac{1}{24}\\y=40\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=24\\y=40\end{matrix}\right.\left(tm\right)\)

Vậy: ...

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{\dfrac{1}{4};1\right\}\end{matrix}\right.\)

\(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\)

\(=\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}-\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt[]{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\dfrac{x-1}{2x+\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}\)

\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(x-2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt[]{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(x-2\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)

a)Xét △HCA và △ACBB

có\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{AHC\left(=90\right)\left(gt\right)}\\\widehat{ACB}chung\end{matrix}\right.\)

⇒△HCA và △ACB (g.g)

b)Có △AHC vuông tại H, HE là đường cao (gt)

⇒EH²=AE.EC ( nhận xét hai △ đồng dạng trong △vuông)

a: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\x+y+2\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-y-3x+y=4-5\\3x-y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x=-1\\y=3x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3\cdot\dfrac{-1}{2}-5=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y-x+y=0-2\\x-y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0y=-2\\x-y=0\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)

a: \(\dfrac{x}{y}+\dfrac{y}{x}>=2\cdot\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)

b: \(\dfrac{1}{x}+\dfrac{1}{y}>=\dfrac{4}{x+y}\)

=>\(\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)

=>\(\left(x+y\right)^2>=4xy\)

=>\(x^2+2xy+y^2-4xy>=0\)

=>\(x^2-2xy+y^2>=0\)

=>\(\left(x-y\right)^2>=0\)(luôn đúng)

a: \(x^2+y^2>=2xy\)

=>\(x^2-2xy+y^2>=0\)

=>\(\left(x-y\right)^2>=0\)(luôn đúng)

b: \(x^2+4xy>=-4y^2\)

=>\(x^2+4xy+4y^2>=0\)

=>\(\left(x+2y\right)^2>=0\)(luôn đúng)

c: \(2\left(x^2+y^2\right)>=\left(x+y\right)^2\)

=>\(2x^2+2y^2-x^2-2xy-y^2>=0\)

=>\(x^2-2xy+y^2>=0\)

=>\(\left(x-y\right)^2>=0\)(luôn đúng)

a: 2a+3>2b+3

=>2a>2b

=>a>b

b: -3a-1>=-3b-1

=>\(-3a>=-3b\)

=>3a<=3b

=>a<=b

c: 5-2a<5-a-b

=>5-2a+a<5-b

=>5-a<5-b

=>a-5>b-5

=>a>b

a: 2<3

=>\(2+8,5\cdot6< 3+8,5\cdot6\)

b: 2<3

=>\(\sqrt{2}< \sqrt{3}\)

=>\(-\sqrt{2}>-\sqrt{3}\)

=>\(30-\sqrt{2}>30-\sqrt{3}\)

c:

Vì 3>2

nên \(3\sqrt{3}>3\sqrt{2}\)

=>\(-3\sqrt{3}< -3\sqrt{2}\)

mà 35<36

nên \(35-3\sqrt{3}< 36-3\sqrt{2}\)

Bài 4:

a: 4x-y=1

=>y=4x-1

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=4x-1\end{matrix}\right.\)

b: x+3y=-2

=>x=-3y-2

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}y\in R\\x=-3y-2\end{matrix}\right.\)

Bài 5:

x+2y-3=0

=>2y=-x+3

=>\(y=\dfrac{-x+3}{2}\)

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=\dfrac{-x+3}{2}\end{matrix}\right.\)

Biểu diễn tập nghiệm: 

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