\(a,5,2x+7\dfrac{2}{5}=6\dfrac{3}{4}\\ \Rightarrow\dfrac{26}{5}x+\dfrac{37}{5}=\dfrac{27}{4}\\ \Rightarrow\dfrac{26}{5}x=-\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{8}\\ b,2,4:\left(\dfrac{-1}{2}-x\right)=1\dfrac{3}{5}\\ \Rightarrow\dfrac{12}{5}:\left(\dfrac{-1}{2}-x\right)=\dfrac{8}{5}\\ \Rightarrow\dfrac{-1}{2}-x=\dfrac{3}{2}\\ \Rightarrow x=-2\)

sửa câu a

\(5,2.x+7\frac{2}{5}=6\frac{3}{4}=>\frac{26}{5}.x+\frac{37}{5}=\frac{27}{4}=>\frac{26}{5}.x=\frac{27}{4}-\frac{37}{5}\)

\(\frac{26}{5}.x=-\frac{13}{20}=>x=-\frac{13}{20}:\frac{26}{5}=>x=-\frac{1}{8}\)

vậy \(x=-\frac{1}{8}\)

a) \(5,2\cdot x+7\frac{2}{5}=6\frac{3}{4}\)

=> \(5,2\cdot x=6\frac{3}{4}-7\frac{2}{5}=\frac{27}{4}-\frac{37}{5}=-\frac{13}{20}\)

=> \(x=-\frac{13}{20}:5,2=-\frac{13}{20}:\frac{26}{5}=-\frac{13}{20}\cdot\frac{5}{26}=-\frac{1}{8}\)

b) \(2,4:\left(\frac{-1}{2}-x\right)=1\frac{3}{5}\)

=> \(\frac{-1}{2}-x=2,4:1\frac{3}{5}=\frac{12}{5}:\frac{8}{5}=\frac{12}{5}\cdot\frac{5}{8}=\frac{3}{2}\)

=> \(x=-\frac{1}{2}-\frac{3}{2}=-\frac{4}{2}=-2\)

\(5,2.x+7\frac{5}{6}=6\frac{3}{4}=>\frac{26}{5}.x+\frac{47}{6}=\frac{27}{4}=>\frac{26}{4}.x=\frac{27}{4}-\frac{47}{6}\)

\(\frac{26}{4}.x=-\frac{13}{12}=>x=-\frac{13}{12}:\frac{26}{4}=>x=-\frac{1}{6}\)

vậy \(x=-\frac{1}{6}\)

\(2,4:\left(-\frac{1}{2}-x\right)=1\frac{3}{5}=>\frac{12}{5}:\left(-\frac{1}{2}-x\right)=\frac{8}{5}=>\left(-\frac{1}{2}-x\right)=\frac{12}{5}:\frac{8}{5}\)

\(\left(-\frac{1}{2}-x\right)=\frac{3}{2}=>x=-\frac{1}{2}-\frac{3}{2}=>x=-2\)

vậy \(x=-2\)

a) \(\dfrac{3}{4}\times x:\dfrac{1}{2}=\dfrac{4}{5}\)

\(\dfrac{3}{4}\times x=\dfrac{4}{5}\times\dfrac{1}{2}\)

\(\dfrac{3}{4}\times x=\dfrac{2}{5}\)

\(x=\dfrac{2}{5}:\dfrac{3}{4}\)

\(x=\dfrac{8}{15}\)

b) \(\left(14-x:6\right)\times7=84\)

\(14-x:6=84:7\)

\(14-x:6=12\)

\(x:6=14-12\)

\(x:6=2\)

\(x=6\times2=12\)

c) \(x\times1,5:2,4=0,3\)

\(x\times1,5=0,3\times2,4\)

\(x\times1,5=0,72\)

\(x=0,72:1,5\)

\(x=0,48\)

d) Thiếu x

Câu 1:

a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$

b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)

c.

\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)

d.

\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)

Câu 2:

a. $x\times \frac{2}{7}=\frac{6}{11}$

$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$

b. $x: \frac{3}{2}=\frac{1}{4}$

$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$

Câu 3:

Coi số lớn là 5 phần thì số bé là 3 phần. Hiệu số phần bằng nhau là:
$5-3=2$ (phần)

Số lớn là: $36:2\times 5=90$

Số bé là: $36:2\times 3=54$

Bài 1 :

A = \(\left(\frac{-2}{3}+\frac{5}{4}-\frac{1}{6}\right)-\frac{24}{10}=\frac{-2}{3}+\frac{5}{4}-\frac{1}{6}+\frac{-24}{10}\)\(=\frac{7}{12}+\frac{-1}{6}+\frac{-12}{5}=\frac{5}{12}+\frac{-12}{5}=\frac{-119}{60}\)

B = \(\frac{13}{15}\cdot0,25\cdot3+\left(\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right)\)\(=\frac{13}{60}\cdot3+\left(\frac{8}{15}-\frac{158}{235}\right)=\frac{13}{20}+\frac{-98}{705}=\frac{1441}{2820}\)

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a) \(5,2x+7\frac{2}{5}=6\frac{3}{4}\)                                                        b) \(2,4:\left(\frac{-1}{2}-x\right)=1\frac{3}{5}\)

\(5,2x+\frac{37}{5}=\frac{27}{4}\)                                                                  \(2,4:\left(\frac{-1}{2}-x\right)=\frac{8}{5}\)

\(5,2x\)            = \(\frac{27}{4}-\frac{37}{5}\)                                                                   \(\frac{-1}{2}-x\)= \(2,4:\frac{8}{5}\)

\(5,2x\)            = \(\frac{-13}{20}\)                                                                               \(\frac{-1}{2}-x\)= \(\frac{3}{2}\)

       x                 = \(\frac{-13}{20}:5,2\)                                                                                      x    = \(\frac{-1}{2}-\frac{3}{2}\)

       x                 = -0,125                                                                                                    x    = \(\frac{-4}{2}\)= -2

Dùng máy tính

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Tính nháp

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

Mn giúp mh với đg cần gấp lắm ln🥺🥺