x=98

x=-105--15=-90

Đầu tiên chúng ta phải đổi các chữ số cùng 1 cột:
abc =>aaa

 ab => bb

  a=> c

Do bb+c<110 nên ta có:

Ta có:874\(\ge\)aaa>874-110=764 =>aaa=777

Ta có:97\(\ge\)bb>97-10=87 => bb=88

Vậy ta có:

c=97-88=9.

=> abc+ab+c=789+78+9=874.

Lời giải:
\(|x-2019|-|x-1|=0\)

\(\Leftrightarrow |x-2019|=|x-1|\)

\(\Rightarrow \left[\begin{matrix} x-2019=x-1\\ x-2019=-(x-1)=1-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 2019=1(\text{vô lý})\\ x=1010\end{matrix}\right.\)

Vậy $x=1010$

Ta có :

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

\(\Rightarrow x.3=5.2\)

\(x.3=10\)

\(x=10:3\)

\(x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\) là giá trị cần tìm

Ta có :\(\dfrac{x}{5}=\dfrac{2}{3}\)

\(\Rightarrow x.3=2.5\)

\(x.3=10\)

\(x=10:3\)

\(\Rightarrow x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\)

Ta có: \(\dfrac{x}{5}=\dfrac{2}{3}\)

=> \(3\times x=2\times5\)

\(3\times x=10\)

\(x=10\div3\)

\(x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\)

Áp dụng BĐT Cô-si:

\(x^3+x^3+8\ge3\sqrt[3]{8x^6}=6x^2\)

\(y^6+y^6+1+1+1+1\ge6\sqrt[6]{y^{12}}=6y^2\)

Cộng vế:

\(2\left(x^3+y^6\right)+12\ge6\left(x^2+y^2\right)\ge30\)

\(\Rightarrow x^3+y^6\ge9\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;1\right)\)

Ta có: \(\frac{x}{1,75}\)=\(\frac{y}{0,75}\) và x-y = 4

⇒ \(\frac{x}{1,75}\)=\(\frac{y}{0,75}\)=\(\frac{x-y}{1,75-0,75}\)=\(\frac{4}{1}\)=4

⇒\(\frac{x}{1,75}\)=4 ⇔ x = 4 . 1,75 = 7

⇒\(\frac{y}{0,75}\)= 4 ⇔ y = 4 . 0,75 = 3

Vậy x = 7 ; y = 3

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1) \(2^x-15=17\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

2) \(\left(7x-11\right)^3=25\cdot5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=825\)

\(\Leftrightarrow7x-11=\sqrt[3]{825}\)

\(\Leftrightarrow7x=11+\sqrt[3]{825}\)

\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)

3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)

\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

4) \(4x+5\left(x+3\right)=105\)

\(\Leftrightarrow9x+15=105\)

\(\Leftrightarrow9x=90\)

\(\Rightarrow x=10\)

5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)

\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)

\(\Leftrightarrow3x+4=34\)

\(\Leftrightarrow3x=30\)

\(\Rightarrow x=10\)

cảm ơn bạn nhá NGUYỄN MINH ĐĂNG