giup minh tra loi nha

1.x=1;5

2.x=11

3.x=1;y=4

4.a)a=2;12        b)a=1;2

nho h cho minh nha

1a. x=-0,8
b)-1va 5/27-(3x-7/9)³=-24/27 mik ko hỉu đề
2.n= 6

 \(A=2018-\left|x-7\right|-\left|y+2\right|\)

Ta có: \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\\left|y+2\right|\ge0\forall y\end{cases}}\Rightarrow2018-\left|x-7\right|-\left|y+2\right|\le2018\)

\(A=2018\Leftrightarrow\hept{\begin{cases}\left|x-7\right|=0\\\left|y+2\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}}\)

Vậy \(A_{m\text{ax}}=2018\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}\)

Tham khảo~

a: -x+5/9=-1/3

nên x=5/9+1/3=5/9+3/9=8/9

b: \(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{11}{15}< x< \dfrac{1}{2}+\dfrac{13}{12}+\dfrac{1}{3}\)

=>31/30<x<23/12

mà x là số nguyên

nên \(x\in\varnothing\)

c: x+5/3=1/81

nên x=1/81-135/81=-134/81

a) 2-(x+3) = 1+2+3+...+99

1+2+3+...+99 → có 99 số hạng

2-(x+3) = (1+99).99 : 2 

2-(x+3) = 4950

x+3 = 2 + 4950 

x+3 = 4952

x = 4952 - 3

x = 4949

b) (x+1)+(x+2)+...+(x+100) = 5750

→ có 100 cặp

(x+x+x+...+x) + ( 1+2+3+...+100 ) = 5750

=> 100x + 5050 = 5750

100x = 5750 - 5050

100x = 700

x = 700 : 100

x = 7

 0o0 Nguyễn Đoàn Tuyết Vy 0o0 bà kêu tui học tốt có nghĩa là học giốt đúng ko

1)

s₁ = 499500

s₂ = 1011010

s₃ = 250901

cho a bang 963+2493+351+x voi x € n tim dieu kien cua x de a chia het cho 9 de a khong chia het cho 9

\(a,\left(7x-11\right)^3=2^5.5^2+200.\)

\(\left(7x+11\right)^3=32.25+200.\)

\(\left(7x+11\right)^3=800+200.\)

\(\left(7x-11\right)^3=1000.\)

\(\left(7x-11\right)^3=10^3.\)

\(\Rightarrow7x-11=10.\)

\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)

Vậy.....

\(b,3^x+25=26.2^2+2.3^0.\)

\(3^x+25=26.4+2.\)

\(3^x+25=104+2.\)

\(3^x+25=106.\)

\(3^x=106-25.\)

\(3^x=81.\)

\(3^x=3^4\Rightarrow x=4\in Z.\)

Vậy.....

\(c,2^x+3.2=64.\)(có vấn đề).

\(d,5^{x+1}+5^x=750.\)

\(5^x.5^1+5^x+1=750.\)

\(5^x\left(5^1+1\right)=750.\)

\(5^x\left(5+1\right)=750.\)

\(5^x.6=750.\)

\(5^x=750:6.\)

\(5^x=125.\)

\(5^x=5^3\Rightarrow x=3\in Z.\)

Vậy.....

\(e,x^{15}=x.\)

\(\Rightarrow x\left(x^{14}-1\right)=0\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right..\)

\(f,\left(x-5\right)^4=\left(x-5\right)^6.\)

\(\Leftrightarrow\left(x-5\right)^4-\left(x-5^6\right)=0.\)

\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0.\)

\(\Leftrightarrow\left(x-5\right)^4\left(1-x+5\right)\left(1+x-5\right)=0.\)

\(\Leftrightarrow\left(x-5\right)^4\left(6-x\right)\left(x-4\right)=0.\)

\(\Leftrightarrow\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\in Z.\)

\(6-x=0\Rightarrow x=6\in Z.\)

\(x-4=0\Rightarrow x=4\in Z.\)

Vậy.....