huhuhuhu help me cứi tui

nguyen dang khoi :viet cai j z troi

Ta có: x + (-23) = (-100) + 77

x + (-23) = - (100 – 77)

x + (-23) = -23

x + (-23) = 0 +(-23)

Nên x = 0

12² < x < 200

144 < x < 200

=> x \(\in\){ 145 ; 146 ; ... ; 199 }

Vậy x \(\in\){ 145 ; 146 ; ... ; 199 }

\(12^2< x< 200\)

\(12^2< x< \sqrt{200}\)

\(12^2< x< \left(14,142...\right)^2\)

\(\Rightarrow x=13^2\)

vậy \(x=13\)

x thuộc {145;146;...199}

Câu 1: Ta có: \(B=3^{75}=\left(3^3\right)^{25}=27^{25}\)

                     \(C=2^{100}=\left(2^4\right)^{25}=16^{25}\)

Vì 27 > 16 nên 27²⁵>16²⁵

Vậy B > C

Câu 2: \(\left\{\left[\left(50-x\right):2+80\right]-5^2.2\right\}.4=200\)

\(\Leftrightarrow\left[\left(50-x\right):2+80-50\right].4=200\)

\(\Leftrightarrow\left(50-x\right):2+30=50\)

\(\Leftrightarrow\left(50-x\right):2=20\)

\(\Leftrightarrow50-x=40\)

\(\Leftrightarrow x=10\)

Vậy x = 10

\(3^{75}=\left(3^3\right)^{25}=27^{25}\)

\(2^{100}=\left(2^4\right)^{25}=8^{25}\)

Vì \(27^{25}>8^{25}\)

\(\Rightarrow3^{75}>2^{100}\)

Câu 1 :\(B=3^{75}=\left(3^3\right)^{25}=27^{25}\)

\(C=2^{100}=\left(2^4\right)^{25}=16^{25}\)

\(\Rightarrow B>C\)

Câu 2 :

\(\left\{\left[\left(50-x\right):2+80\right]-5^2.2\right\}.4=200\)

\(\Rightarrow\left[\left(50-x\right):2+80\right]=200:4:2:5^2=1\)

\(\Rightarrow\left(50-x\right):2+80=1\)

\(\Rightarrow x=50+79.2\)

\(\Rightarrow x=..........\)

Nếu x \(\ge\) 0

x + 1 + x + 2+ ....... +x + 100 = 101x

100x +  100 x 101  : 2 = 101x

x = 100 x 101 : 2

x = 5050

Nếu x < 0 

-(x  + 1) - (x + 2)  -.... - (x + 100) = 101x

-x - 1 - x - 2 - .... - x - 100 = 101x

- 100x - (100 x 101 : 2) = 101x

-100x - 5050=  101x

101x + 100x = -5050

201x = -5050

x = -5050/201 (loại , vì không phải số nguyên)

Vậy x = 5050

Nếu x $\ge$≥ 0

x + 1 + x + 2+ ....... +x + 100 = 101x

100x +  100 x 101  : 2 = 101x

x = 100 x 101 : 2

x = 5050

Nếu x < 0 

-(x  + 1) - (x + 2)  -.... - (x + 100) = 101x

-x - 1 - x - 2 - .... - x - 100 = 101x

- 100x - (100 x 101 : 2) = 101x

-100x - 5050=  101x

101x + 100x = -5050

201x = -5050

x = -5050/201 (loại , vì không phải số nguyên)

Vậy x = 5050

\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)

\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)

\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)

\(2a=3b\text{⇒}a=\dfrac{3b}{2}\) , \(5b=7c\text{⇒}c=\dfrac{5c}{7}\) 

\(3a-7b+5c\) \(=-30\) 

⇔ \(3.\dfrac{3b}{2}-7b+5.\dfrac{5b}{7}=-30\)  

⇔\(63b-98b+50b=-420\) 

⇔\(b=-28\) ⇒\(\left\{{}\begin{matrix}a=-42\\c=-20\end{matrix}\right.\)

=> x - 100 = 300 

=> x = 300 - 100

=> x = 200

    x-100=100+200

\(\Rightarrow\)x-100=300

\(\Rightarrow\)x       =300-100

\(\Rightarrow\)x       =200

k cho minh nha

x - 100 = 100 + 200 

x - 100 = 300

x = 300 + 100 

x = 400