Bài 1:

Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Leftrightarrow2x=\frac{1440}{144}=10\)

\(\Rightarrow x=5\)

Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)

=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)

a) Ta có: \(x^2< 5x\)

         \(\Leftrightarrow x^2-5x< 0\)

         \(\Leftrightarrow x.\left(x-5\right)< 0\)

    Ta có bảng xét dấu:

x x x-5 h 0 5 - ​​ - - - + + + + +

\(\Rightarrow\)\(x< 0\)hoac \(x>5\)

b) Để \(\frac{x+3}{2-x}\le0\)

    Ta có bảng xét dấu:

    x x-3 2-x Thương 2 3 + + - - - - + - -

\(\Rightarrow\)\(\orbr{\begin{cases}x\le2\\x\ge3\end{cases}}\)

\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\)

=> \(\left(\frac{12}{25}\right)^x=\frac{9}{25}-\frac{81}{625}\)

=> \(\left(\frac{12}{25}\right)^x=\frac{225}{625}-\frac{81}{625}=\frac{144}{625}\)

=> \(\left(\frac{12}{25}\right)^x=\left(\frac{12}{25}\right)^2\)

Vậy x = 2

                 Bài làm :

Ta có :

\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\)

 \(\Leftrightarrow\left(\frac{12}{25}\right)^x=\frac{9}{25}-\frac{81}{625}\)

 \(\Leftrightarrow\left(\frac{12}{25}\right)^x=\frac{225}{625}-\frac{81}{625}=\frac{144}{625}\)

 \(\Leftrightarrow\left(\frac{12}{25}\right)^x=\left(\frac{12}{25}\right)^2\)

\(\Leftrightarrow x=2\)

Vậy x=2

Vì \(2x=3y\)\(\Rightarrow\)\(\frac{x}{3}=\frac{y}{2}\)\(\Rightarrow\)\(\frac{x}{15}=\frac{y}{10}\)

     \(3y=5z\)\(\Rightarrow\)\(\frac{y}{5}=\frac{z}{3}\)\(\Rightarrow\)\(\frac{y}{10}=\frac{z}{6}\)

     \(\Rightarrow\)\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

Đặt \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=k\) \(\left(k\ge0\right)\)

       \(\Rightarrow\)\(\hept{\begin{cases}x=15k\\y=10k\\z=6k\end{cases}}\)

Ta có: \(x-y+z=-33\)

    \(\Leftrightarrow15k-10k+6k=-33\)

    \(\Leftrightarrow11k=-33\)

    \(\Leftrightarrow k=-3\)

     \(\Rightarrow\hept{\begin{cases}x=15.\left(-3\right)=-45\\y=10.\left(-3\right)=-30\\z=6.\left(-3\right)=-18\end{cases}}\)

Vậy \(x=-45,\)\(y=-30,\)\(z=-18\)

Theo đề bài , ta có : 2x=3y =5z 

=> \(\frac{2x}{30}\)\(=\frac{3y}{30}\)\(=\frac{5z}{30}\)

\(\Leftrightarrow\frac{x}{15}\) \(=\frac{y}{10}\) \(=\frac{z}{6}\)

A/D tính chất dãy tỉ số bằng nhau , ta có : 

x/15 = y/10 = z/6 =  x -y+z / 15 -10 + 6 = -33/11 = -3

Suy ra :

x/15 = -3 <=> x = -45

y/10 = -3 <=> y =  -30

z/6 = -3 <=> z = -18

Vậy , x-..;y=...;z=....

a) Ta có : \(2.21=3.14\)\(\Rightarrow\)lập được các tỉ lệ thức là : \(\frac{2}{3}=\frac{14}{21};\frac{2}{14}=\frac{3}{21};\frac{3}{2}=\frac{21}{14};\frac{14}{2}=\frac{21}{3}\)

b) Ta có : \(-0,24.1,61=-0,46.0,84\)\(\Rightarrow\)lập được các tỉ lệ thức là : \(\frac{-0,24}{-0,46}=\frac{0,84}{1,61};\frac{-0,24}{0,84}=\frac{-0,46}{1,61};\frac{-0,46}{-0,24}=\frac{1,16}{0,84};\frac{0,84}{-0,24}=\frac{1,16}{-0,46}\)

\(\frac{3^6.45^5-15^{13}.5^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3^2.5\right)^5-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}=\frac{3^{16}.5^5-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}\)

\(\frac{3^{13}.5^4\left(3^3.5-1\right)}{2.3^{12}.5^6}=\frac{3.44}{2.5^2}=\frac{3.22}{5^2}=\frac{66}{25}\)

Tìm x

\(\left(152\frac{2}{4}-148\frac{3}{8}\right):0,2=x:0,3\)

\(x:0,3=\left(152\frac{1}{2}-148\frac{3}{8}\right):0,2\)

\(x:0,3=\left(152\frac{4}{8}-148\frac{3}{8}\right):0,2\)

\(x:0,3=4\frac{1}{8}:0,2\)

\(x:\frac{3}{10}=\frac{33}{8}:\frac{1}{5}\)

\(x:\frac{3}{10}=\frac{165}{8}\)

\(x=\frac{165}{8}.\frac{3}{10}\)

\(x=\frac{99}{16}\)

Vậy \(x=\frac{99}{16}\).

\(\frac{x^2}{6}=\frac{24}{25}\)

\(\Rightarrow25x^2=24.6=144\)

\(\Rightarrow x^2=\frac{144}{25}\)

\(\Rightarrow x=\frac{12}{5}\)

Vậy \(x=\frac{12}{5}\).

\(\frac{x-1}{x+5}=\frac{6}{7}\)

\(\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(7x-7=6x+30\)

\(7x-6x=7+30\)

x = 37

Vậy x = 18.

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)

\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)

x² - 2x + 7x - 14 = x² - x + 4x - 4

(x² - x²) - (2x - 7x - x + 4x) = 14 - 4

- (2x - 7x - x + 4x) = 10

- (-2x) = 10

2x = 10

x = 5

Vậy x = 5.

\(5x=7y\Leftrightarrow\frac{x}{7}=\frac{y}{5}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có :

\(\frac{y}{5}=\frac{x}{7}=\frac{y-x}{5-7}=\frac{18}{-2}=-9\)

Suy ra :

+) \(\frac{y}{5}=-9\Leftrightarrow y=-45\)

+) \(\frac{x}{7}=-9\Leftrightarrow x=-63\)

Theo đề bài , ta có : 5x=7y  <=> x/7 = y/5 và y-x = 18

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

x/7 = y/5 = y-x/5-7 = 18/ -2 = -9

Suy ra :

x/7 = -9 <=> x = -63

y/5 = -9 <=> y = -45

Vậy , x= - 63 ; y = -45

ĐK \(2018x\ge0\Rightarrow x\ge0\)

Khi đó \(x+\frac{1}{2018}\ge0;x+\frac{2}{2018}\ge0;...;x+\frac{2017}{2018}\ge0\)

Ta có \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)(Vế trái có 2017 hạng tử)

<=> \(x+\frac{1}{2018}+x+\frac{2}{2018}+...+x+\frac{2017}{2018}=2018x\)

<=> \(\left(x+x+...x\right)+\left(\frac{1}{2018}+\frac{2}{2018}+...+\frac{2017}{2018}\right)=2018x\)

           2017 hạng tử x                   2017 số hạng

=> \(2017x+\frac{1+2+...+2017}{2018}=2018x\)

=> \(x=\frac{2017.\left(2017+1\right):2}{2018}\)

\(\Rightarrow x=\frac{2017}{2}=1008,5\)(tm)

Vậy x = 1008,5

Vì \(\left|x+\frac{1}{2018}\right|\ge0\forall x\)

    \(\left|x+\frac{2}{2018}\right|\ge0\forall x\)

    \(\left|x+\frac{3}{2018}\right|\ge0\forall x\)

     .......................................

    \(\left|x+\frac{2017}{2018}\right|\ge0\forall x\)

\(\Rightarrow\)\(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|\ge0\forall x\)

mà \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)

 \(\Rightarrow\)\(2018x\ge0\forall x\)\(\Rightarrow\)\(x\ge0\)

 \(\Rightarrow\)\(x+\frac{1}{2018}+x+\frac{2}{2018}+x+\frac{3}{2018}+...+x+\frac{2017}{2018}=2018x\)

\(\Leftrightarrow\)\(2017x+\frac{1}{2018}+\frac{2}{2018}+\frac{3}{2018}+...+\frac{2017}{2018}=2018x\)

\(\Leftrightarrow\)\(\frac{1+2+3+...+2017}{2018}=x\)

\(\Leftrightarrow\)\(x=\frac{\left[\left(2017+1\right).2017\right]:2}{2018}\)

\(\Leftrightarrow\)\(x=\frac{2035153}{2018}\)

\(\Leftrightarrow\)\(x=\frac{2017}{2}=1008,5\)

Vậy \(x=1008,5\)

a. 2/3 = 14/21, 3/2 = 21/14, 3/21 = 2/14, 21/3 = 14/2

b. Tương tự với a. nhé