\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)

\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)

\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)

\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)

Vậy.....

\(b,\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3.\)

\(=4-1\dfrac{5}{27}.\left(-\dfrac{27}{8}\right).\)

\(=4-\dfrac{32}{27}.\left(-\dfrac{27}{8}\right).\)

\(=4-\left(-4\right).\)

\(=4+4=8.\)

Vậy.....

\(c,\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}.\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}+0+0+...+0-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}-\dfrac{1}{100}.\)

\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}.\)

Vậy.....

a) x - 3/97 + x - 2/98 = x - 1/99 + x/100

<=> x + 1/99 + 1 + x + 2/98 + 1 + x + 3/97 + 1 + (x + 4/96 + 1 + x + 5/95 + 1 + x + 10/90 + 1) = 0

<=> x + 100/99 + x + 100/98 + x + 100/97 + (x + 100/96 + x + 100/95 + x + 100/90) = 0

<=> (x + 100)(1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90) = 0

Mà 1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90 khác 0

=> x + 100 = 0

=> x = -100

c) (1/1.2 + 1/2.3 + ... + 1/99.100) - 2x = 1/2

<=> (1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100) - 2x = 1/2

<=> (1 - 1/100) - 2x = 1/2

<=> 99/100 - 2x = 1/2

<=> -2x = 1/2 - 99/100

<=> -2x = -49/100

<=> x = 49/200

=> x = 49/200

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

3]

c]S=1.2+2.3+3.4+...+99.100

  3S=1.2.3+2.3.[4-1]+3.4.[5-2]+...+99.100.[101-98]

  3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.10-98.99.100

  3S=99.100.101

Con lai ban tu giai nha

Bài 1: 

a) Ta có: \(x\left(x^2-4\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;2;-2\right\}\)

b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)

\(\Leftrightarrow2x-3-3x-x+5=40\)

\(\Leftrightarrow-2x+2=40\)

\(\Leftrightarrow-2x=38\)

hay x=-19

Vậy: x=-19

Bài 2: 

a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)

\(=-45\cdot\left(12+34+54\right)\)

\(=-45\cdot100\)

\(=-4500\)

b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)

\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)

\(=43\cdot57-33\cdot57\)

\(=57\cdot\left(43-33\right)\)

\(=57\cdot10=570\)

\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)

\(\Leftrightarrow8x^2+49x-15=0\)

\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)