Bài 7:

a) \(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)=0+0+0+0=0\)

b) \(=\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{3}{11}-\dfrac{3}{11}\right)+\left(\dfrac{3}{17}-\dfrac{3}{17}\right)+\dfrac{3}{19}=0+0+0+\dfrac{3}{19}=\dfrac{3}{19}\)

Bài 8:

1 giờ vòi 1 chảy được là: \(1:6=\dfrac{1}{6}\left(phần\right)\)

1 giờ vòi 2 chảy đc là: \(1:5=\dfrac{1}{5}\left(phần\right)\)

1 giờ vòi 3 chảy được là: \(1:4=\dfrac{1}{4}\left(phần\right)\)

1 giờ 3 vòi chảy được: \(\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{4}=\dfrac{37}{60}\left(phần\right)\)

Thể tích của vật là:

\(V=\dfrac{m}{D}=\dfrac{210}{10500}=0,02\left(m^3\right)\)

Lực đẩy Ác si mét tác dụng lên vật là:

\(F_A=d.V=10000.0,02=200\left(N\right)\)

a) \(2\left(x+3\right)-4=2x-5\)

\(\Leftrightarrow2x+6-4=2x-5\Leftrightarrow2=-5\left(VLý\right)\Rightarrowđpcm\)

b) \(2\left(1-4x\right)-7=-8x\)

\(\Leftrightarrow2-8x-7=-8x\Leftrightarrow-5=0\left(VLý\right)\Rightarrowđpcm\)

1) ĐKXĐ: \(x\ne1\)

\(pt\Leftrightarrow\dfrac{2x-1+x-1-1}{x-1}=0\Leftrightarrow3x-3=0\Leftrightarrow3x=3\Leftrightarrow x=1\left(ktm\right)\)

Vậy \(S=\varnothing\)

2) ĐKXĐ: \(x\ne7\)

\(pt\Leftrightarrow\dfrac{x-8-8\left(x-7\right)+1}{x-7}=0\Leftrightarrow-7x+49=0\Leftrightarrow x=7\left(ktm\right)\)

Vậy \(S=\varnothing\)

3) ĐKXĐ: \(x\ne2\)

\(pt\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\Leftrightarrow4x-8=0\Leftrightarrow x=2\left(ktm\right)\Rightarrow S=\varnothing\)

4) ĐKXĐ: \(x\ne2\)

\(pt\Leftrightarrow\dfrac{1+3\left(x-2\right)+x-3}{x-2}=0\Leftrightarrow4x-8=0\Leftrightarrow x=2\left(ktm\right)\Rightarrow S=\varnothing\)

5) ĐKXĐ: \(x\ne-1\)

\(pt\Leftrightarrow\dfrac{5x+2x+2+12}{2\left(x+1\right)}=0\Leftrightarrow7x+14=0\Leftrightarrow x=-2\left(tm\right)\)

6) ĐKXĐ: \(x\ne0,x\ne2\)

\(pt\Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)-2}{x\left(x-2\right)}=0\Leftrightarrow x^2+x+2=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(VLý\right)\Rightarrow S=\varnothing\)

ĐKXĐ: \(x\ne y,x\ne z,y\ne z\)

\(\dfrac{4}{\left(y-x\right)\left(z-x\right)}+\dfrac{1}{\left(y-x\right)\left(y-z\right)}+\dfrac{3}{\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{4\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}-\dfrac{x-z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\dfrac{3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{4\left(y-z\right)-\left(x-z\right)+3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\dfrac{2x+y-3z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{\sqrt{x}+1}\left(đk:x\ne1,x\ge0\right)\)

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

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

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

a) \(x^2-5x\ge0\Leftrightarrow x\left(x-5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le0\end{matrix}\right.\)

b) Thiếu vế phải rồi ạ

c) Do \(x^2+1\ge1>0\forall x\)

\(\Rightarrow-x^2+3x+4\le0\Leftrightarrow-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{25}{4}\le0\)

\(\Leftrightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le0\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2\ge\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}\ge\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}\le-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{3+\sqrt{5}}{2}\\x\le\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

a) Rút gọn được \(\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

c) \(H=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\Rightarrow H^2=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}\)

\(\Rightarrow H^2-H=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}-\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}=\dfrac{xy-\sqrt{xy}\left(x-\sqrt{xy}+y\right)}{\left(x-\sqrt{xy}+y\right)^2}\)

\(=\dfrac{2xy-x\sqrt{xy}-y\sqrt{xy}}{\left(x-\sqrt{xy}+y\right)^2}=\dfrac{-\sqrt{xy}\left(x-2\sqrt{xy}+y\right)}{\left(x-\sqrt{xy}+y\right)^2}=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\)

Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(x-\sqrt{xy}+y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow H^2-H=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\le0\Rightarrow H^2\le H\)

Mà \(H\ge0\left(cmt\right)\Rightarrow H\le\sqrt{H}\)