\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+.....+\left|x+\dfrac{100}{101}\right|=101x\left(1\right)\)

VT(1) \(\ge0\) \(\Rightarrow VP\left(1\right)\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=100x+\dfrac{5050}{101}=101x\\ \Rightarrow x=50\)

tim gtln

Ta có: \(\left|x+\frac{1}{101}\right|\ge0\) \(\forall x\)

\(\left|x+\frac{2}{101}\right|\ge0\) \(\forall x\)

\(\left|x+\frac{3}{101}\right|\ge0\) \(\forall x\)

\(............\)

\(\left|x+\frac{100}{101}\right|\ge0\) ∀\(x\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\) \(\forall x\)

\(\Leftrightarrow101x\ge0\)

\(\Leftrightarrow x\ge0\)

\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)

100 hạng tử x 100 số hạng

\(\Leftrightarrow100x+\left(\frac{\left(100+1\right)\cdot100:2}{101}\right)=101x\)

\(\Leftrightarrow100x+\frac{101\cdot50}{101}=101x\)

\(\Leftrightarrow50=101x-100x\)

\(\Rightarrow x=50\)

Ta có: \(\left|x+\frac{1}{101}\right|\ge0\); \(\left|x+\frac{2}{101}\right|\) \(\ge0\); ...; \(\left|x+\frac{100}{101}\right|\ge0\)

\(\Rightarrow101x\ge0\)

và \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\); \(\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\); ...; \(\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)

Thay vào đề bài ta đc:

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)

\(\Rightarrow\) \(100x\) + \(\left(\frac{1+2+...+101}{101}\right)=101x\)

\(\Rightarrow100x+101=101x\)

\(\Rightarrow x=101\)

Vậy \(x=101.\)

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x (1)

điều kiện:101x\(\ge\) 0 \(\Rightarrow\) x\(\ge\) 0

từ (1) \(\Rightarrow\) \(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}\)=101x

\(\Rightarrow\) 100x+(\(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\))=101x

\(\Rightarrow\) 100x+\(\frac{5050}{101}\)=101x

\(\Rightarrow\) \(\frac{5050}{101}\)=101x-100x

\(\Rightarrow\) x=50

k bt mk lm sai hay lm đúng nữa

nếu mk lm sai thì thôi nha!

\(!x+\frac{1}{101}!+!x+\frac{2}{101}!+...+!x+\frac{100}{101}!=101x\) (1)

VT tổng các số không âm => VT>=0 vậy \(VP\ge0\Rightarrow x\ge0\)

với x>=0 biểu thức trong GT tuyệt đối >0 => bỏ dấu trị tuyệt đối biểu thức không đối

do vậy ta có (1) \(\Leftrightarrow\left(x+\frac{1}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow100.x+\left(\frac{1}{101}+...+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow x=\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}=\frac{1}{101}\left(1+2+...+100\right)=\frac{1}{101}\left(\frac{100.101}{2}\right)=50\)

đáp số: x=50

Nhận xét :

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)

\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)

b) 

Tương tự câu a) , phương trình tương đương với : 

\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)

\(\Rightarrow x=\frac{97}{195}\)

c) 

Tương tự câu a) , phương trình tương đương với : 

\(99x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)

\(\Rightarrow x=\frac{99}{100}\)

Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{1}{102}\right|+....+\left|x+\frac{100}{101}\right|>0\)

\(\Rightarrow101x>0\)

\(\Rightarrow x>0\)

\(\Rightarrow\left(x+\frac{1}{101}\right)+.....+\left(x+\frac{100}{101}\right)=101x\)

\(\Rightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)

\(\Rightarrow x=\frac{\left(100+1\right)100:2}{101}\)

\(\Rightarrow x=\frac{50.101}{101}\)

\(\Rightarrow x=50\)

Vậy x = 50

Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)

=> \(101x\ge0\)

=> \(x\ge0\)

=> \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)

            100 số x                          100 phân số

=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)

=> \(\frac{101.50}{101}=101x-100x\)

=> \(x=50\)

Vì \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\forall x\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\forall x\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(100x+\frac{1+2+...+100}{101}=101x\)

\(101x-100x=\frac{5050}{101}\)

\(x=50\)

Vậy x = 50

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)

\(KĐ:101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

\(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)

\(100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)

\(\Rightarrow101-100x=\frac{1+2+....+100}{101}\)

\(x=\frac{\left(1+100\right)\left(100-1+1\right):2}{101}\)

\(x=\frac{101.100:2}{101}\)

\(x=50\)

\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=101x\)

Ta có : \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{101}\right|\ge0\\\left|x+\dfrac{1}{102}\right|\ge0\\....\\\left|x+\dfrac{100}{101}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{101}\right|=x+\dfrac{1}{101}\\\left|x+\dfrac{2}{101}\right|=x+\dfrac{2}{101}\\....\\\left|x+\dfrac{100}{101}\right|=x+\dfrac{100}{101}\end{matrix}\right.\)

\(\Rightarrow x+\dfrac{1}{101}+x+\dfrac{2}{101}+x+\dfrac{3}{101}+...+x+\dfrac{100}{101}=101x\)

\(\Rightarrow100x+\dfrac{1+2+3+...+100}{101}=101x\)

\(\Rightarrow100x+\dfrac{5050}{101}=101x\)

\(\Rightarrow100x+50=101x\)

\(\Rightarrow101x-100x=50\)

\(\Rightarrow x=50\)

Vậy \(x=50\)

a) ( x² - 1 )( x - 101 ) + 101x( x + 1 ) = 101

<=> x³ - 101x² - x + 101 + 101x² + 101x - 101 = 0

<=> x³ + 100x = 0

<=> x( x² + 100 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x^2+100=0\end{cases}}\Leftrightarrow x=0\)( vì x² + 100 ≥ 100 > 0 ∀ x )

b) x⁴ - 3x²( 2x - 3 ) = 0

<=> x⁴ - 6x³ + 9x² = 0

<=> x²( x² - 6x + 9 ) = 0

<=> x²( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x^2=0\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

a,\(\left(x^2-1\right)\left(x-101\right)+101x\left(x+1\right)=101\)

\(\Leftrightarrow x^3-101x^2-x+101+101x^2+101x=101\)

\(\Leftrightarrow x^3+100x=101-101\)

\(\Leftrightarrow x^3+101x=0\)

\(\Leftrightarrow x\left(x^2+101\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x^2+101\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-101\end{cases}\Rightarrow}x=0}\)