\(3x+60=-120\)

\(3x=-120-60\)

\(3x=-180\)

\(x=-180\div3\)

\(x=-60\)

\(3x+60=-120\\ \Rightarrow3x=-180\\ \Rightarrow x=-60.\) 

Vậy ....

`#3107.101107`

`(2x + 1)^3 = 343`

`\Rightarrow (2x + 1)^3 = 7^3`

`\Rightarrow 2x + 1 = 7`

`\Rightarrow 2x = 6`

`\Rightarrow x = 3`

Vậy, `x = 3.`

x=3

Bài khó lắmmm tr oiii

B = \(\dfrac{4}{7}\) = \(\dfrac{4.4}{7.4}\) = \(\dfrac{16}{28}\); I = \(\dfrac{6}{13}\) = \(\dfrac{6.\left(-2\right)}{13.\left(-2\right)}\) = \(\dfrac{-12}{-26}\)

N = \(\dfrac{-5}{13}\) = \(\dfrac{-5.3}{13.3}\) = \(\dfrac{-15}{39}\); T = \(\dfrac{7}{21}\) = \(\dfrac{7.4}{21.4}\) = \(\dfrac{28}{84}\)

U = \(\dfrac{4}{11}\) = \(\dfrac{4.5}{11.5}\) = \(\dfrac{20}{55}\);   O = \(\dfrac{5}{25}\) = \(\dfrac{5.3}{25.3}\) = \(\dfrac{15}{75}\)

H = \(\dfrac{1}{5}\) = \(\dfrac{1.11}{5.11}\) = \(\dfrac{11}{15}\);    A = \(\dfrac{5}{8}\) = \(\dfrac{5.5}{8.5}\) = \(\dfrac{25}{40}\)

G =  \(\dfrac{-3}{17}\) = \(\dfrac{-3.5}{17.5}\) = \(\dfrac{-15}{85}\); D = \(\dfrac{4}{16}\) = \(\dfrac{4.5}{16.5}\) = \(\dfrac{20}{80}\)

B. $\frac{4}{7}=\frac{16}{28}$

N. $\frac{-5}{13}=\frac{-15}{39}$

U. $\frac{4}{11}=\frac{20}{55}$

H. $\frac{1}{5}=\frac{11}{55}$

G. $\frac{-3}{17}=\frac{-15}{85}$

I. $\frac{6}{13}=\frac{-12}{-26}$

T. $\frac{7}{21}=\frac{28}{84}$

O. $\frac{5}{25}=\frac{15}{75}$

A. $\frac{5}{8}=\frac{25}{40}$

D. $\frac{4}{16}=\frac{20}{80}$

Ta có: \(xy-2x+3y=-5\)

\(\Rightarrow\left(xy-2x\right)+3y-6=-5-6\)

\(\Rightarrow x\left(y-2\right)+3\left(y-2\right)=-11\)

\(\Rightarrow\left(x+3\right)\left(y-2\right)=-11\)

Vì \(x,y\) nguyên nên \(x+3;y-2\) có giá trị nguyên

\(\Rightarrow x+3;y-2\) là các ước của \(-11\)

Ta có bảng sau:

Vì \(x,y\) nguyên nên ta được các cặp giá trị \(\left(x;y\right)\) là:

\(\left(-2;-9\right);\left(8;1\right);\left(-4;13\right);\left(14;3\right)\)

\(Toru\)

(-8).82+(-8).38+8.20

= (-8).82+(-8).38+(-8).(-20)

= (-8).(82+38-20)

= (-8).100 = -800

Chữ không có tâm đối xứng là của S.

Chọn C.

chu nao do nha moi nguoi

   -36 - 12

= -(36 + 12)

= - 48

Chọn D. - 48 

 -36 - 12

= -(36 + 12)

= - 48

a) Ta có: \(\left|x+5\right|\ge0\forall x\)

\(\Rightarrow\left|x+5\right|+2023\ge2023\forall x\)

\(\Rightarrow A\ge2023\forall x\)

Dấu \("="\) xảy ra khi: \(x+5=0\Leftrightarrow x=-5\)

Vậy \(Min_A=2023\) khi \(x=-5\).

b) Ta có: \(\left\{{}\begin{matrix}\left|2x+6\right|\ge0\forall x\\\left|y+3x\right|\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|\ge0\forall x,y\)

\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|+25\ge25\forall x,y\)

\(\Rightarrow B\ge25\forall x,y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}2x+6=0\\y+3x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-6\\y=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-6:2=-3\\y=-3\cdot\left(-3\right)=9\end{matrix}\right.\)

Vậy \(Min_B=25\) khi \(x=-3;y=9\).

c) Ta có: \(\left\{{}\begin{matrix}\left|12-3x\right|\ge0\forall x\\\left|-y-4x\right|\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|\ge0\forall x,y\)

\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|-12\ge-12\forall x,y\)

\(\Rightarrow C\ge-12\forall x,y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}12-3x=0\\-y-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=12\\y=-4x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=12:3=4\\y=-4\cdot4=-16\end{matrix}\right.\)

Vậy \(Min_C=-12\) khi \(x=4;y=-16\).

\(\mathit{Toru}\)

Theo đề bài,ta có:

\(49⋮x;56⋮x\) và \(x\) lớn nhất \(\Rightarrow x\inƯCLN\left(49;56\right)=7\)

Vậy \(x=7\)

Theo đề : x thuộc UCLN(49,56)

Mà : 49 = 7^2 và 56 = 2^3 . 7

=> UCLN (49,56) = 7

=> x = 7