\(A=1\frac{3}{4}x^4-\frac{1}{8}x^3-1\frac{1}{4}x^2+\frac{2}{5}x+\frac{4}{7}\)

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Bài 3: 

 \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

6 tháng 3 2020

1. A = 75(42004 + 42003 +...+ 4+ 4 + 1) + 25

    A = 25 . [3 . (42004 + 42003 +...+ 4+ 4 + 1) + 1]

    A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 4+ 3 . 4 + 3 + 1)

    A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 4+ 3 . 4 + 4)

    A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100

6 tháng 3 2020

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

\(A=\frac{3}{4}-1+1=\frac{3}{4}\)

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

7/4.x+3/2=-4/5

7/4.x=-4/5-3/2

7/4.x=-23/10

x=-23/10:7/4

x=-46/35

vậy x=-46/35

1/4+3/4.x=3/4

1.x=3/4

x=3/4:1

x=3/4

vậy x=3/4

x.(1/4+1/5)-(1/7+1/8)=0

x.9/20-15/56=0

x.51/280=0

x=0:51/280

x=0

vậy x=0

3/35-(3/5+x)=2/7

(3/5+x)=3/35-2/7

(3/35+x)=-1/5

x=-1/5-3/5

x=-4/5

vậy x=-4/5

\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)

\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)

\(\frac{7}{4}.x=\frac{-7}{10}\)

\(x=\frac{-7}{10}:\frac{7}{4}\)

\(x=\frac{-2}{5}\)

\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)

\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)

\(\frac{3}{4}.x=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{3}{4}\)

\(x=\frac{2}{3}\)

\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)

\(x.\frac{9}{20}-\frac{15}{56}=0\)

\(x.\frac{9}{20}=\frac{15}{56}\)

\(x=\frac{15}{56}:\frac{9}{20}\)

\(x=\frac{25}{42}\)

\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)

\(\frac{3}{5}+x=\frac{-1}{5}\)

\(x=\frac{-1}{5}-\frac{3}{5}\)

\(x=\frac{-4}{5}\)

Học tốt

Bài 1:

a) Ta có: \(\frac{-5}{8}+x=\frac{4}{9}\)

\(\Leftrightarrow x=\frac{4}{9}-\frac{-5}{8}=\frac{32}{72}-\frac{-45}{72}\)

hay \(x=\frac{77}{72}\)

Vậy: \(x=\frac{77}{72}\)

b) Ta có: \(1\frac{3}{4}\cdot x+1\frac{1}{2}=-\frac{4}{5}\)

\(\Leftrightarrow\frac{7}{4}\cdot x+\frac{3}{2}=-\frac{4}{5}\)

\(\Leftrightarrow\frac{7}{4}\cdot x=-\frac{4}{5}-\frac{3}{2}=-\frac{23}{10}\)

\(\Leftrightarrow x=\frac{-23}{10}:\frac{7}{4}=\frac{-23}{10}\cdot\frac{4}{7}\)

hay \(x=-\frac{46}{35}\)

Vậy: \(x=-\frac{46}{35}\)

c) Ta có: \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)

\(\Leftrightarrow\frac{3}{4}x=\frac{2}{4}\)

\(\Leftrightarrow x=\frac{2}{4}:\frac{3}{4}=\frac{2}{4}\cdot\frac{4}{3}\)

hay \(x=\frac{2}{3}\)

Vậy: \(x=\frac{2}{3}\)

d) Ta có: \(x\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)

\(\Leftrightarrow x\cdot\frac{9}{20}-\frac{15}{56}=0\)

\(\Leftrightarrow x\cdot\frac{9}{20}=\frac{15}{56}\)

\(\Leftrightarrow x=\frac{15}{56}:\frac{9}{20}=\frac{15}{56}\cdot\frac{20}{9}\)

hay \(x=\frac{25}{42}\)

Vậy: \(x=\frac{25}{42}\)

e) Ta có: \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\Leftrightarrow\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)

\(\Leftrightarrow\frac{-18}{35}-x=\frac{2}{7}\)

\(\Leftrightarrow-x=\frac{2}{7}-\frac{-18}{35}=\frac{2}{7}+\frac{18}{35}=\frac{4}{5}\)

hay \(x=-\frac{4}{5}\)

Vậy: \(x=-\frac{4}{5}\)

f) Ta có: \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)

\(\Leftrightarrow\frac{1}{7}\cdot\frac{1}{x}=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)

\(\Leftrightarrow\frac{1}{x}=\frac{-3}{14}:\frac{1}{7}=-\frac{3}{14}\cdot7=-\frac{3}{2}\)

\(\Leftrightarrow x=\frac{1\cdot2}{-3}=\frac{2}{-3}=-\frac{2}{3}\)

Vậy: \(x=-\frac{2}{3}\)

g) Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)