a) Vì x và y tỉ lệ thuận vs nhau 
\(=>\dfrac{x_2}{x_1}=\dfrac{y_2}{y_1}=>x_1=\dfrac{x_2.y_1}{y_2}=\dfrac{2.\left(\dfrac{-3}{4}\right)}{\dfrac{1}{7}}=\dfrac{-21}{2}\)
b) \(\dfrac{y_1}{x_1}=\dfrac{y_2}{x_2}=\dfrac{3}{-4}=\dfrac{-3}{4}=>y_1=\dfrac{-3}{4}x_1\)
Mà \(y_1-x_1=-2\)
\(< =>\dfrac{-3}{4}x_1-x_1=-2\)
\(< =>\dfrac{7}{4}x_1=2\)
\(< =>x_1=\dfrac{8}{7}\)
\(=>y_1=\dfrac{-6}{7}\)

a) \(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(=\left(sin^2x-1+sin^2x\right).1=2sin^2x-1\)
b) \(\dfrac{1}{sin^2x}+\dfrac{1}{cos^2x}=\dfrac{cos^2x+sin^2x}{sin^2x}+\dfrac{cos^2x+sin^2x}{cos^2x}\)
\(=1+\dfrac{cos^2x}{sin^2x}+1+\dfrac{sin^2x}{cos^2x}=2+tan^2x+cot^2x\)

\(-3,75+\left(\dfrac{39}{10}-6,25\right)-\left(-10,1\right)\)
\(=\dfrac{-15}{4}-\dfrac{47}{20}-\left(-10,1\right)=4\)

cái dề :)))

Số gạo đó ăn đc trong số ngày là
200 x 45 : 150 = 60 ngày

\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2}\right).\dfrac{\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}.\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\left(x-\sqrt{x}\right)-2\sqrt{x}+2-x-\sqrt{x}-2\sqrt{x}.\dfrac{\left(\sqrt{x}+1\right)}{2}\)
\(=-6\sqrt{x}.\dfrac{\left(\sqrt{x}+1\right)}{2}=-3x-3\sqrt{x}\)
 

Đề 3 - Bài 1
\(7+\dfrac{7}{12}-\dfrac{1}{2}+3-\dfrac{1}{12}-5=\dfrac{7}{12}-\dfrac{1}{12}+7+3-5-\dfrac{1}{2}\)
\(=\dfrac{6}{12}+5-\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+5=5\)
Bài 2 
Ta có 
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{6-4-3}{12}=\dfrac{-1}{12}\)
\(\dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)=\dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}=\dfrac{1-3+8}{48}=\dfrac{1}{8}\)
Vậy \(\dfrac{-1}{12}< x< \dfrac{1}{8}\) (vì \(x\in Z=>x=0\))
Đề 4 - Bài 1
a) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}=>\dfrac{17}{6}-x+\dfrac{7}{6}=\dfrac{7}{4}\)
\(=>-x=\dfrac{7}{4}-\dfrac{17}{6}-\dfrac{7}{6}=>-x=\dfrac{-54}{24}=>x=\dfrac{9}{4}\)
b) \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}=>\dfrac{3}{35}-\dfrac{3}{5}+x=\dfrac{2}{7}\)
\(=>x=\dfrac{2}{7}-\dfrac{3}{35}+\dfrac{3}{5}=>x=\dfrac{28}{35}=>x=\dfrac{4}{5}\)
Bài 2 
Ta có 
\(\dfrac{3}{4}-\dfrac{5}{6}=\dfrac{-1}{12}\)
\(1-\left(\dfrac{2}{3}-\dfrac{1}{4}\right)=1-\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
Vậy \(\dfrac{-1}{12}\le\dfrac{x}{12}< \dfrac{7}{12}\)
Vì \(x\in Z=>x\in\left\{-1;0;1;2;3;4;5;6\right\}\)

Vận tốc người đi xe đạp đi đc nửa quãng đường còn lại là
\(V_{tb}=\dfrac{2}{\dfrac{1}{v_1}+\dfrac{1}{v_2}}=\dfrac{2}{\dfrac{1}{4}}=8\)
\(=>V_{tb}=\dfrac{2}{\dfrac{1}{12}+\dfrac{1}{v_2}}=8=>v_2=6\) km/h