a) \(-x-\dfrac{3}{7}=-\dfrac{7}{12}\Leftrightarrow-x=-\dfrac{13}{84}\Leftrightarrow x=\dfrac{13}{84}\)

b) \(\dfrac{6}{5}-\left|x-1\right|=0,3\Leftrightarrow-\left|x-1\right|=\dfrac{3}{10}-\dfrac{6}{5}\Leftrightarrow\left|x-1\right|=\dfrac{9}{10}\)

TH1: \(x-1=\dfrac{9}{10}\Leftrightarrow x=\dfrac{19}{10}\)

TH2: \(x-1=-\dfrac{9}{10}\Leftrightarrow x=\dfrac{1}{10}\)

Ta có: Trước đây mua 5kg gạo phải trả 60.000 đồng

\(\rightarrow\) Mỗi kg gạo trước đây có giá 60.000:5 = 12.000 (đồng)

mà  hiện nay giá bán mỗi kg gạo giảm 2000 đồng

\(\rightarrow\) Mỗi kg gạo trước đây có giá 12.000-2000=10.000 (đồng)

\(\Rightarrow\) 60.000 đồng hiện nay có thể mua: 60.000:10.000 = 6 (kg gạo)

Kẻ đường cao AH

Ta có: AD=AB(=x) \(\Rightarrow\Delta ABD\) cân tại A

\(\Rightarrow\) AH cũng là đường trung trực

\(\Rightarrow\) DH=HB=9m

\(\Rightarrow CH=AD+DH=7+9=16\left(m\right)\)

Xét \(\Delta ABC\) vuông tại A, đường cao AH có:

\(AH^2=BH\cdot CH\) (HTL)

\(\Rightarrow AH=\sqrt{9\cdot16}=12\) (m)

Xét \(\Delta AHB\) vuông tại H có:

\(AB^2=AH^2+HB^2\) (ĐL Py-ta-go)

\(\Rightarrow AB=\sqrt{12^2+9^2}=15\) (m)

\(\Rightarrow x=15\left(m\right)\)

a)

\(A=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) \(A=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)

\(\Leftrightarrow2x-5\sqrt{x}+2=0\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)(T/m)

c) \(\dfrac{1}{A}>\dfrac{3\sqrt{x}}{4}\)

\(\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)

\(\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}}-\sqrt{x}>0\)

\(\Leftrightarrow\dfrac{-\sqrt{x}+1}{\sqrt{x}}>0\)

\(\Leftrightarrow-\sqrt{x}+1>0\left(\sqrt{x}>0\right)\)

\(\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

mà \(x\ge0\) mà x dưới mẫu \(\Rightarrow0< x< 1\)

*Rút gọn N

\(N=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{5\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(N=\dfrac{x-\sqrt{x}-5\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(N=\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(Q=M\cdot N=\dfrac{x-\sqrt{x}+3}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{\sqrt{x}}\)

Xét hiệu Q-2 ta có: \(Q-2=\dfrac{x-\sqrt{x}+3}{\sqrt{x}}-2=\dfrac{x-3\sqrt{x}+3}{\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}}{\sqrt{x}}>0\)

\(\Rightarrow Q-2>0\Rightarrow Q>2\)

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

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

\(M=\dfrac{x}{\sqrt{x}-1}\)

b) \(M=-\dfrac{1}{2}\Leftrightarrow\dfrac{x}{\sqrt{x}-1}=-\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=-2x\Leftrightarrow2x+\sqrt{x}-1=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\Leftrightarrow2\sqrt{x}-1=0\Leftrightarrow x=\dfrac{1}{4}\)

c) \(M>1\Leftrightarrow\dfrac{x}{\sqrt{x}-1}>1\)

\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

Ta có: \(x-\sqrt{x}+1=x-2\cdot\dfrac{1}{2}\cdot\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{4}\right)^2+\dfrac{3}{4}>0\)

\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

\(\left(x+\dfrac{2}{3}\right)-\dfrac{1}{4}=0\Leftrightarrow x+\dfrac{2}{3}=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{5}{12}\)

a) Xét \(\Delta\)AHB vuông tại H có đường cao HD có:

\(AH^2=AD\cdot AB\) (HTL) (1)

Xét \(\Delta\)AHC vuông tại H có đường cao HE có:

\(AH^2=AE\cdot AC\) (HTL) (2)

Từ (1)(2) \(\Rightarrow AD\cdot AB=AE\cdot AC\)  (đpcm)

b) Ta có: \(AD\cdot AB=AE\cdot AC\) (theo a)

\(\rightarrow\dfrac{AD}{AE}=\dfrac{AC}{AB}\)

Xét \(\Delta ADE\) và \(\Delta ACB\) có:

\(\widehat{A}\) chung

\(\dfrac{AD}{AE}=\dfrac{AC}{AB}\)

\(\rightarrow\Delta ADE\sim\Delta ACB\left(cgc\right)\)

\(\Rightarrow\widehat{ADE}=\widehat{ACB}\) (đpcm)

c) Xét \(\Delta\)ABC vuông tại A đường cao AH có:

\(AB^2=HB\cdot BC\) (HTL)

\(AC^2=HC\cdot BC\) (HTL)

\(\Rightarrow\dfrac{AB^2}{AC^2}=\dfrac{HB\cdot BC}{HC\cdot BC}=\dfrac{HB}{HC}\) (đpcm)

a) Ta có: \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=\sqrt{\left(3y\right)^2}=\left|3y\right|\)

Vì y>0 nên ta có: \(\left|3y\right|=3y\)

b) Ta có: \(\dfrac{\sqrt{16x^4y^6}}{\sqrt{128x^6y^6}}=\sqrt{\dfrac{16x^4y^6}{128x^6y^6}}=\sqrt{\dfrac{1}{8x^2}}=\sqrt{\left(\dfrac{1}{2\sqrt{2}x}\right)^2}=\left|\dfrac{1}{2\sqrt{2}x}\right|\)

Vì x<0 và \(y\ne0\) nên ta có: \(\left|\dfrac{1}{2\sqrt{2}x}\right|=-\dfrac{1}{2\sqrt{2}x}\)