\(\left(x+1\right)^{200}=\left(2x-3\right)^{200}\)

\(\Rightarrow2x-x=1+3\)

\(\Rightarrow x=4\)

Vậy x = 4

\(\left(x+1\right)^{200}=\left(2x-3\right)^{200}\)

\(\Rightarrow x+1=2x-3\)

\(\Rightarrow x-2x=-3-1\)

\(\Rightarrow-x=-4\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

(x+1)²⁰⁰= (2x-3)²⁰⁰

=> x+1=2x-3

=> x-2x= 1-3

=> -x= -2

=> x=2

(x + 1)²⁰⁰ = (2x - 3 )²⁰⁰

x + 1 = 2x - 3

x + 1 - 2x = -3

-x + 1 = -3

-x = -3 - 1

-x = -4

x = 4

Vậy x = 4

Vì: \(\left(x-12+y\right)^{200}\ge0;\left(x-4-y\right)^{200}\ge0\)

=> \(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}\le0\)

\(\Leftrightarrow\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)

\(\Leftrightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)\(\Leftrightarrow\begin{cases}x+y=12\\x-y=4\end{cases}\)\(\Leftrightarrow\begin{cases}x=8\\y=4\end{cases}\)

bài này trong đề thi hsg lớp  7 đúng ko

Ta có \(\left|x+1\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;...;\)\(\left|x+\frac{1}{190}\right|\ge0\)    \(\forall x\)

=> \(\left|x+1\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{190}\right|\ge0\)   \(\forall x\)

=> \(20x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\)  =>  \(x+1>0,x+\frac{1}{3}>0,x+\frac{1}{6}>0,...,x+\frac{1}{190}>0\)

                        => \(\left|x+1\right|=x+1,\left|x+\frac{1}{3}\right|=x+\frac{1}{3},\left|x+\frac{1}{6}\right|=x+\frac{1}{6},...,\left|x+\frac{1}{190}\right|=x+\frac{1}{190}\)

=> \(x+1+x+\frac{1}{3}+x+\frac{1}{6}+...+x+\frac{1}{190}=20x\)

=> \(19x+\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)=20x\)

=> \(x=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)\)

Gọi \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)

=> \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

=> \(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

=> \(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

=> \(\frac{1}{2}A=1-\frac{1}{20}\)

=> \(A=\frac{19}{10}\)

Thay vào ta có 

=> \(x=-\frac{19}{10}\)

mk nhầm nha bạn \(x=\frac{19}{10}\)

`Answer:`

Vì `N=40<=>b=40-(3+2+3+3+8+9+2+1)=9`

Theo đề cho:

`6,25=\frac{a.3+3.2+4.3+5.3+6.8+7.9+8.9+9.2+10.1}{40}`

`<=>6,25=\frac{3a+6+12+15+48+63+72+18+10}{40}`

`<=>6,25=\frac{3a+244}{40}`

`<=>3a+244=40.6,25`

`<=>3a+244=250`

`<=>a=2`

OLM.VN

a) Ta có \(\sqrt{17}>\sqrt{16}=4\)

\(\sqrt{26}>\sqrt{25}=5\)

Khi đó \(\sqrt{17}+\sqrt{26}+1>4+5+1=10\)    (1)

Mà \(\sqrt{99}< \sqrt{100}=10\)                                            (2)

Từ (1) và (2) suy ra \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

Vậy....

Cảm mơn bn rất nhiều <3 <3 <3