a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)

hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)

b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow x+100=0\)

hay x=-100

Mn nhớ giải chi tiết ra nha

a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

2A=2/2x4+2/4x6+..+2/98x100

=> 2A=1-1/4+1/4-1/6+..+1/98-1/100

=> 2A=1-1/100

=> 2A=99/100

=> A=99/100:2

=> A=99/200

\(A=1\cdot2^2+2\cdot3^2+3\cdot4^2+\cdots+98\cdot99^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+4^2\left(4-1\right)+\cdots+99^2\left(99-1\right)\)

\(=\left(2^3+3^3+\cdots+99^3\right)-\left(2^2+3^2+\cdots+99^2\right)\)

\(=\left(1^3+2^3+\cdots+99^3\right)-\left(1^2+2^2+3^2+\cdots+99^2\right)\)

\(=\left(1+2+3+\cdots+99\right)^2-\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}\)

\(=\left(99\cdot\frac{100}{2}\right)^2-\frac{99\cdot100\cdot199}{6}\)

\(=\left(99\cdot50\right)^2-33\cdot50\cdot199=\left(3\cdot33\cdot50\right)^2-33\cdot50\cdot199\)

\(=9\cdot33^2\cdot50^2-33\cdot50\cdot199=33\cdot50\left(9\cdot33\cdot50-199\right)\)

\(=33\cdot50\cdot14651=24174150\)

       \(\frac{2x+1}{4-x}=\frac{x-1}{5}\)

\(\Rightarrow\)\(5.\left(2x+1\right)=\left(4-x\right).\left(x-1\right)\)

\(\Rightarrow\)\(10x+5=2x-4\)

\(\Rightarrow\)\(10x-2x=-4-5\)

\(\Rightarrow\)\(8x=-9\)

\(\Rightarrow\)\(x=-\frac{9}{8}\)

323400 nha !

A = 1.2 + 2.3 + 3.4 + ... + 98.99 

3A = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 98.99.3

3A = 1.2.3 + 2.3.(4 -1) + 3.4.(5 - 2) + ...+ 98.99.(100 - 97)

3A = (1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100) - (0.1.2 + 1.2.3  + 2.3.4 + ... + 97.98.99)

3A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100 

3A = 98.99.100

=> A = \(\frac{98.99.100}{3}=323400\)

câu 1 

=> x+1/2+x+1/3+x+1/4-x-1/5-x-1/6=0

=> (x+x+x-x-x)+(1/2+1/3+1/4-1/5-1/6)=0

=> x+43/60=0

=> x = -43/60

câu dưới làm tương tự bạn nhé!