x+x-1+x-2+x-3+x-4+...+x-50 = 225

<=> 51x-(1+2+...+50) = 225

<=> 51x - 1275 = 225

<=> 51x = 1500

<=> x = 500/17

Thanks

\(x+x-1+x-2+...+x-50=225\)

=> \(\left(x+x+x+...+x\right)-\left(1+2+...+50\right)=225\) ( có 51 hạng tử x)

=> \(51x-\left(1+2+...+50\right)=225\)   (*)

Xét \(1+2+...+50\)

Có \(\left(50-1\right)+1=50\)  hạng tử

=> \(1+2+...+50= \left(50+1\right).50 :2 = 1275\)

Thay vào (*) : \(51x-1275=225\)

=> \(x=\frac{500}{17}\)

x+x-1+x-2+...+X-50=225

=> (x+x+...+x)-(1+2+3+...+50)=225

=> 51x-1275=225

=> 51x=1500

=> x=30

Vậy x=30

a: \(x^4-2x^3-25x^2+50x=0\)

=>\(x^3\left(x-2\right)-25x\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x^3-25x\right)=0\)

=>x(x-2)(x^2-25)=0

=>x(x-2)(x+5)(x-5)=0

=>x∈{0;2;-5;5}

b: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>x∈{1;2}

c: \(9x^2-4-2\left(3x-2\right)^2=0\)

=>(3x-2)(3x+2)-(3x-2)(6x-4)=0

=>(3x-2)(3x+2-6x+4)=0

=>(3x-2)(-3x+6)=0

=>(x-2)(3x-2)=0

=>x∈{2;2/3}

d: \(9x^2+90x+225-\left(x-7\right)^2=0\)

=>\(\left(3x+15\right)^2-\left(x-7\right)^2=0\)

=>(3x+15-x+7)(3x+15+x-7)=0

=>(2x+22)(4x+8)=0

=>2(x+11)*4*(x+2)=0

=>(x+11)(x+2)=0

=>x∈{-11;-2}

e: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+3x+5\right)=0\)

mà \(x^2+3x+5=x^2+3x+\frac94+\frac{11}{4}=\left(x+\frac32\right)^2+\frac{11}{4}>0\forall x\)

nên x-2=0

=>x=2

g: (x+1)(x+2)(x+3)(x+4)-24=0

=>\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

=>\(\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24=0\)

=>\(\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

mà \(x^2+5x+10=x^2+5x+\frac{25}{4}+\frac{75}{4}=\left(x+\frac52\right)^2+\frac{75}{4}\ge\frac{75}{4}>0\forall x\)

nên \(x^2+5x=0\)

=>x(x+5)=0

=>x∈{0;-5}

Bài 1:a)  |x - 3| = 2x + 4

=> \(\orbr{\begin{cases}x-3=2x+4\\x-3=-2x-4\end{cases}}\)

=> \(\orbr{\begin{cases}x-2x=4+3\\x+2x=-4+3\end{cases}}\)

=> \(\orbr{\begin{cases}-x=7\\3x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=-\frac{1}{3}\end{cases}}\)

Vậy ...

b) Để M có giá trị nguyên thì 2n - 7 \(⋮\)n - 5 

   <=> 2(n - 5) + 3 \(⋮\)n - 5 

   <=> 3 \(⋮\)n - 5

  <=> n - 5 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng : 

Vậy ...

cảm ơn bạn nhiều Kuruba Kaito

b, (x - 1)(x - 2)= 0

Suy ra , ta có 2 trường hợp

Th1:(x -1)(x - 2)=0

Suy ra (x-1)=0 suy ra x=1

Th2:(x -1)(x - 2)=0

Suy ra (x -2)=0 suy ra x =2

Vậy x thuộc {1;2}

TÍCH NHÉ BẠN

\(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\)

\(\dfrac{z}{5}=\dfrac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x^2+y^2}{9+16}=\dfrac{x^2+y^2}{25}=\dfrac{225}{25}=9\)

\(\Rightarrow x=\sqrt{9\cdot9}=9\)

\(\Rightarrow y=\sqrt{9\cdot16}=12\)

\(\Rightarrow z=\sqrt{9\cdot25}=15\)

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

\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{225}{25}=9\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=9.9=81\\y^2=16.9=144\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=9\\y=12\end{matrix}\right.\)

\(\Rightarrow z=\dfrac{9}{3}.5=15\)

Vậy \(\left\{{}\begin{matrix}x=9\\y=12\\z=15\end{matrix}\right.\) thỏa đề bài

\(4^x=64\)

\(\Rightarrow x=3\)

\(15^x=225\)

\(\Rightarrow x=2\)

\(3^x:9=27\)

\(\Leftrightarrow3^x=243\)

\(\Leftrightarrow x=5\)

\(x^{2018}=0\)

\(\Leftrightarrow x=0\)

\(x^{50}=x\)

\(\Rightarrow x\in\left\{0;1\right\}\)

\(300^x=1\)

\(\Rightarrow x=0\)

`\(12^x=144\)

\(\Rightarrow x=2\)

4^x=64=4³=> x=3

15^x=225=15²=> x=2

3^x :9 = 27

3^x=3³x3²=3⁵=> x=5

X²⁰¹⁸=0=> x=0

X⁵⁰=x=> x=1 hoặc 0

300^x=1=> x=0

12^x=144=12²=> x=2

(x-1+1)(x+1):2=210

x(x+1):2=210

x(x+1)=420=20.21

=>x=20