32 - 42 * (-19) + 38 * 5

= 32 - (-798) + 190

= 32 + 798 + 190

= 830 + 190

= 1020

32 - 42 x (-19) + 38 x 5

= 32 - ( - 798 ) + 190

= 830 + 190

= 1020

\(\frac{2^3.3}{2^23^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha 

A = \(\frac{2^3.3}{2^2.3^2.5}\)

A = \(\frac{2^2.3.2}{2^2.3.3.5}\)

A = \(\frac{2}{3.5}\)

A = \(\frac{2}{15}\)

\(-x-\frac{3}{4}=-\frac{8}{11}=>-x=-\frac{8}{11}+\frac{3}{4}=\frac{1}{44}=>x=-\frac{1}{44}\)

-x-3/4=-8/11

=0.02272727....

11/12-(2/5+x)=2/3

=-0.15

a) ta có : \(|-x+8|\ge0\)

=> \(|-x+8|-21\ge-21\)

=> A \(\ge-21\)

Vậy A đạt GTNN là -21 khi x=8

b) ta có :\(|-x-17|+|y-36|\ge0\)

=> \(|-x-17|+|y-36|+12\ge0+12\)

=> B \(\ge12\)

Vậy B đạt GTNN là 12 khi x=-17 và y =36

c) ta có: \(-|2x-8|\le0\)

=> \(-|2x-8|-35\le0-35\)

=>  C \(\le-35\)

Vậy C đạt GTLN là -35 khi 2x-8=0==> x=4

d) ta có : \(3.\left(3x-12\right)^2\ge0\)

=> \(3.\left(3x-12\right)^2-35\ge0-35\)

=>  \(D\ge-35\)

Vậy D  đạt GTNN là -35 khi x =4

e) ta có : \(-3.|2x+50|\le0\)

=>: \(-21-3.|2x+50|\le0-21\)

=> E \(\le-21\)

vậy E đạt GTLN là -21 khi x=-25

\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)

\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)

\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)

\(=>x=0\)

\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)

\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)

\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)

tu do \(=>x=-7,8;...;0;1;2;3;4\)
 

Ảnh hơi mờ 

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)

\(a,\left(x+3\right)\left(y+2\right)=1\)

=> x+3 và y+2 thuộc UC(1)={1; -1}

Vậy x=-2; y=-4

       x=-1; y=-4

Câu sau tương tự

\(a,\left(x+3\right)\left(y+2\right)=1\)

Th1 : \(\hept{\begin{cases}x+3=1\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)

Th2 : \(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)

KL : \(\left\{\left(x=-2;y=-1\right);\left(x=-4;y=-3\right)\right\}\)

\(d,3x+4y-xy=16\)

\(=3x-xy+4y-12=4\)

\(\Rightarrow-x\left(y-3\right)+4\left(y-3\right)=4\)

\(\Rightarrow\left(y-3\right)\left(4-x\right)=4\)

Chia các trường hợp như câu a của chị ra em nhé

a) => 4/3x = 7/9 - 4/9 = 1/3

=> x = 1/3 : 4/3 = 1/4

b) => 5/2 - x = 9/14 : (-4/7) = -9/8

=> x = 5/2 - (-9/8) = 5/2 + 9/8 = 29/8

c) => 3x = 2 và 2/3 - 3/4 = 8/3 - 3/4 = 23/12

=> x = 23/12 : 3 = 23/36

D) => -5/6 - x = 1/4

=> x = -5/6 - 1/4 = -13/12

a) \(\dfrac{4}{9}+\dfrac{4}{3}x=\dfrac{7}{9}\)

\(\dfrac{4}{3}x=\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{4}{3}\)

\(x=\dfrac{1}{4}\)

b) \(\left(\dfrac{5}{2}-x\right)\left(-\dfrac{4}{7}\right)=\dfrac{9}{14}\)

\(\dfrac{5}{2}-x=\dfrac{9}{14}:\left(-\dfrac{4}{7}\right)=-\dfrac{9}{8}\)

\(x=\dfrac{5}{2}-\left(-\dfrac{9}{8}\right)\)

\(x=\dfrac{29}{8}\)

c) \(3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(3x=\dfrac{8}{3}-\dfrac{3}{4}=\dfrac{23}{12}\)

\(x=\dfrac{23}{12}:3\)

\(x=\dfrac{23}{36}\)

d) \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(x=-\dfrac{13}{12}\)

\(\frac{1}{15}\) và \(-7\)

Biết\(-7=\frac{-7}{1}\)nên MSC là : 15

Ta có:

\(\frac{1}{15}=\frac{1.1}{15.1}=\frac{1}{15}\)

\(\frac{-7}{1}=\frac{-7.15}{1.15}=\frac{-105}{15}\)

Vậy quy đồng mẫu số các phân số \(\frac{1}{15}\) và \(-7\) ta được \(\frac{1}{15}\) và \(\frac{-105}{15}\)