Lời giải:
a.

$2\frac{3}{4}-a+\frac{1}{4}=1\frac{1}{2}$

$3-a=1\frac{1}{2}$

$a=3-1\frac{1}{2}=\frac{3}{2}$

b. 

$3\frac{1}{4}-a-1\frac{3}{4}=\frac{7}{8}$

$\frac{3}{2}-a=\frac{7}{8}$

$a=\frac{3}{2}-\frac{7}{8}=\frac{5}{8}$

c.

$2\frac{5}{6}-1\frac{1}{2}-a=\frac{1}{6}$

$a=2\frac{5}{6}-1\frac{1}{2}-\frac{1}{6}=\frac{7}{6}$

a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)

\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)

\(3-a=\dfrac{3}{2}\)

\(a=3-\dfrac{3}{2}\)

\(a=\dfrac{6}{2}-\dfrac{3}{2}\)

\(a=\dfrac{3}{2}\)

b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)

\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)

\(0-a=\dfrac{7}{8}\)

\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)

c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)

\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)

\(\dfrac{8}{6}-a=\dfrac{1}{6}\)

\(a=\dfrac{8}{6}-\dfrac{1}{6}\)

\(a=\dfrac{7}{6}\)

a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)

     a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)

     a  = 2 + 1 - 1 - \(\dfrac{1}{2}\)

     a  = 2 - \(\dfrac{1}{2}\)

     a = \(\dfrac{3}{2}\)

b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)

    (3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)

                     a = - \(\dfrac{7}{8}\)

c,    2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a  = \(\dfrac{1}{6}\)

    a =  2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\)  - \(\dfrac{1}{6}\) 

     a =  (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)

     a = 1 +  \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)

     a = \(\dfrac{7}{6}\)

Số tiền 3 quyển vở mỏng và 18 quyển vở dày là :

\(90000x3=270000\left(đồng\right)\)

Số tiền (18-4)=14 quyển vở dày là :

\(270000-88000=182000\left(đồng\right)\) 

Số tiền 1 quyển vở dày là :

\(182000:14=13000\left(đồng\right)\)

Số tiền 1 quyển vở mỏng là :

\(90000-6x13000=12000\left(đồng\right)\)

Đáp số...

3\(x\) - 5 ⋮ \(x\) + 1 ( đkxđ: \(x\) \(\ne\) -1)

3\(x\) + 3 - 8 ⋮ \(x\) + 1

3.(\(x\) + 1) - 8⋮ \(x\) + 1

                 8 \(⋮\) \(x\) + 1

\(x\in\){ -8; -4; -2; -1; 1; 2; 4; 8}

\(x\) \(\in\){-9; -5; -3; -2; 0; 1; 3; 7}

a) \(x^2+2x+1=\left(x+1\right)^2\)

b) \(x^2+8x+16=\left(x+4\right)^2\)

c) \(x^2+6x+9=\left(x+3\right)^2\)

d) \(4x^2+4x+1=\left(2x+1\right)^2\)

e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)

f) \(4x^2+12x+9=\left(2x+3\right)^2\)

g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)

h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)

a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)²

b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)²

c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)²

d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)²

36 + 4

6 + 4 = 10 viết 0 nhớ 1; 3 nhớ 1 bằng 4 vậy

36 + 4 = 40 

36 + 4 = 40

\(3^{x+1}-3^x=1458\)

\(\Leftrightarrow3^x.3-3^x=1458\)

\(3^x.\left(3-1\right)=1458\)

\(3^x.2=1458\)

\(3^x=1458:2\)

\(3^x=729\)

\(3^x=3^6\)

\(\Rightarrow x=6\)

3^\(x+1\) - 3\(^x\) = 1458

3\(^x\).( 3 - 1) = 1458

3\(^x\). 2 = 1458

3\(^x\)     = 1458: 2

3\(^x\)     = 729

3\(^x\)      = 3⁶

\(x\) = 6

\(\Rightarrow\dfrac{z}{5}=\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z-x+y}{5-3+4}=1\)

\(\Rightarrow x=3;y=4;z=5\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{z-x+y}{3-4+5}=\dfrac{6}{4}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}\cdot3=4,5\)

\(y=\dfrac{3}{2}\cdot4=6\)

\(z=\dfrac{3}{2}\cdot5=7,5\)

Sửa đề:

\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)

\(A=4.\dfrac{33}{68}\)

\(A=\dfrac{33}{17}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)

A = \(\dfrac{11}{17}\)