Bạn gõ đầy đủ đề ra nhé. 

5(2x-3)-4(5x-7)=19-2(x+11)

<=> 10x-15-20x+28=19-2x-22

<=> 10x-20x+2x=19-22+15-28

<=> -8x=-16

<=> x=2

\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=19-22+15-28\)

\(\Leftrightarrow-8x=-16\)

\(\Leftrightarrow x=2\)

Ta có: \(A^2=\frac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}=\frac{20xy-12xy}{20xy+12xy}=\frac{8xy}{32xy}=\frac{1}{4}\)

Vì \(2y< 3x< 0\Rightarrow3x-2y>0,3x+2y< 0\Rightarrow A< 0\)

Vậy A= \(\frac{-1}{2}\)

Ta có :

\(A^2=\frac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}\)\(=\frac{20xy-12xy}{20xy+12xy}\)\(=\frac{8xy}{32xy}\)\(=\frac{1}{4}\)

\(Do\)\(2y< 3x< 0\Rightarrow3x-2y>0;3x+2y< 0\Rightarrow A< 0\)

Vậy \(A=-\frac{1}{2}\)

Ta có: \(x^2-xy-2y^2=0\Leftrightarrow x^2+xy-2xy+2y^2=0\)\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

Vì \(x+y\ne0\Rightarrow x=2y\)

=> \(A=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

\(\frac{\left(7x+1\right)\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\left(x-3\right)}{2}\)

⇔ \(\frac{7x^2-13x-2}{10}+\frac{2}{5}-\frac{x^2-4x+4}{5}-\frac{x^2-4x+3}{2}=0\)

⇔ \(\frac{7x^2-13x-2}{10}+\frac{4}{10}-\frac{2\left(x^2-4x+4\right)}{10}-\frac{5\left(x^2-4x+3\right)}{10}=0\)

⇔ \(\frac{7x^2-13x-2}{10}+\frac{4}{10}-\frac{2x^2-8x+8}{10}-\frac{5x^2-20x+15}{10}=0\)

⇔ \(\frac{7x^2-13x-2+4-2x^2+8x-8-5x^2+20x-15}{10}=0\)

⇔ \(\frac{15x-21}{10}=0\)

⇔ 15x - 21 = 0

⇔ x = 21/15

\(\frac{\left(7x+1\right)\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\left(x-3\right)}{2}\)

\(\Leftrightarrow\frac{7x^2-13x-2}{10}+\frac{4}{10}=\frac{2\left(x^2-4x+4\right)}{10}+\frac{5\left(x^2-4x+3\right)}{10}\)

\(\Leftrightarrow\frac{7x^2-13x-2+4}{10}=\frac{2x^2-8x+8+5x^2-20x+15}{10}\)

\(\Leftrightarrow\frac{7x^2-13x+2}{10}=\frac{7x^2-28x+23}{10}\)

\(\Leftrightarrow\frac{7x^2-13x+2}{10}-\frac{7x^2-28x+23}{10}=0\)

\(\Leftrightarrow7x^2-13x+2-7x^2+28x-23=0\)

\(\Leftrightarrow15x-21=0\)

\(\Leftrightarrow15x=21\)

\(\Leftrightarrow x=\frac{21}{15}=\frac{7}{5}\)

\(P=\frac{x+y}{xyz}=\frac{x}{xyz}+\frac{y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\)

Áp dụng Bunyakovsky dạng phân thức : \(\frac{1}{yz}+\frac{1}{xz}\ge\frac{4}{z\left(x+y\right)}\)(1)

Ta có : \(\sqrt{z\left(x+y\right)}\le\frac{x+y+z}{2}\)( theo AM-GM )

=> \(z\left(x+y\right)\le\left(\frac{x+y+z}{2}\right)^2=\left(\frac{6}{2}\right)^2=9\)

=> \(\frac{1}{z\left(x+y\right)}\ge\frac{1}{9}\)=> \(\frac{4}{z\left(x+y\right)}\ge\frac{4}{9}\)(2)

Từ (1) và (2) => \(P=\frac{x+y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\ge\frac{4}{z\left(x+y\right)}\ge\frac{4}{9}\)

=> P ≥ 4/9

Vậy MinP = 4/9, đạt được khi x = y = 3/2 ; z = 3

\(\frac{5x-1}{3}+\frac{7x-1,1}{3}-\frac{1,5-5x}{7}=\frac{9x-0,7}{4}\)

⇔ \(\frac{5x-1+7x-1,1}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

⇔ \(\frac{12x-2,1}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

⇔ \(\frac{28\left(12x-2,1\right)}{84}-\frac{12\left(1,5-5x\right)}{84}-\frac{21\left(9x-0,7\right)}{84}=0\)

⇔ \(\frac{336x-58,8}{84}-\frac{18-60x}{84}-\frac{189x-14,7}{84}=0\)

⇔ \(\frac{336x-58,8-18+60x-189x+14,7}{84}=0\)

⇔ \(\frac{207x-62,1}{84}=0\)

⇔ 207x - 62, 1 = 0

⇔ 207x = 62, 1

⇔ x = 0, 3

\(\frac{5x-1}{3}+\frac{7x-1.1}{3}-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\left(\frac{5x-1}{3}+\frac{7x-1.1}{3}\right)-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\left(\frac{5x-1+7x-1.1}{3}\right)-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\frac{12x-2.1}{3}-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\frac{28\left(12x-2.1\right)}{84}-\frac{12\left(1.5-5x\right)}{84}-\frac{21\left(8x-0,7\right)}{84}=0\)

\(\Leftrightarrow\frac{336x-58.8-18+60x-189x+14.7}{84}=0\)

\(\Leftrightarrow336x-58.8-18+60x-189x+14.7=0\)

\(\Leftrightarrow207x-62.1=0\)

\(\Leftrightarrow207x=62.1\)

\(\Leftrightarrow x=\frac{62.1}{207}=\frac{3}{10}=0.3\)

\(\frac{7x^2-14x-5}{15}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)

=> \(\frac{7x^2-14x-5}{15}=\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}\)

=> \(\frac{7x^2-14x-5}{15}=\frac{3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)}{15}\)

=> \(\frac{7x^2-14x-5}{15}=\frac{7x^2+22x-2}{15}\)

=> 7x² - 14x - 5 = 7x² + 22x - 2

=> -14x - 5 + 22x - 2

=> 36x = -3

=> x = -1/12

\(\frac{7x^2-14x-5}{15}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)

\(\Leftrightarrow\frac{7x^2-14x-5}{15}=\frac{4x^2+4x+1}{5}-\frac{x^2-2x+1}{3}\)

\(\Leftrightarrow\frac{7x^2-14x-5}{15}-\frac{4x^2+4x+1}{5}+\frac{x^2-2x+1}{3}=0\)

\(\Leftrightarrow\frac{7x^2-14x-5}{15}-\frac{3\left(4x^2+4x+1\right)}{15}+\frac{5\left(x^2-2x+1\right)}{14}=0\)

\(\Leftrightarrow7x^2-14x-5-12x^2-12x-3+5x^2-10x+5=0\)

\(\Leftrightarrow-36x-3=0\)

\(\Leftrightarrow-36x=3\)

\(\Leftrightarrow x=\frac{3}{-36}=-\frac{1}{12}\)