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27 tháng 12 2018

Đáp án C

22 tháng 7 2018

hình như sai đề câu b vs d bn ơi

22 tháng 7 2018

x là nhân ak

30 tháng 4 2017

Bài 1:

a)

\(A=\dfrac{-5}{6}\cdot\dfrac{3}{10}\\ =\dfrac{\left(-5\right)\cdot3}{6\cdot10}\\ =\dfrac{-1}{4}\)

b)

\(B=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12}\\ =\dfrac{4}{12}-\dfrac{3}{12}+\dfrac{1}{12}\\ =\dfrac{4-3+1}{12}\\ =\dfrac{1}{6}\)

30 tháng 4 2017

Bài 2:

\(A=\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{4}{5}-\dfrac{14}{5}\right|:\dfrac{8}{3}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{-10}{5}\right|\cdot\dfrac{3}{8}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+2\cdot\dfrac{3}{8}\\ =\dfrac{-3}{4}+\dfrac{3}{4}\\ =0\)

\(B=\left(\dfrac{-1}{2}\right)^2:1\dfrac{3}{8}+25\%\cdot\dfrac{3}{11}\\ =\left(\dfrac{-1}{2}\right)^2:\dfrac{11}{8}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{1}{4}\cdot\dfrac{8}{11}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{8}{44}+\dfrac{9}{44}\\ =\dfrac{17}{44}\)

\(C=\dfrac{-8}{5}+0,6+\left|\dfrac{-1}{2}\right|+\dfrac{1}{2}\\ =\dfrac{-8}{5}+\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(\dfrac{-8}{5}+\dfrac{3}{8}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =\left(-1\right)+1\\ =0\)

\(D=\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}:\dfrac{13}{11}+1\dfrac{5}{9}\\ =\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}\cdot\dfrac{11}{13}+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot\left(\dfrac{2}{13}+\dfrac{11}{13}\right)+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot1+\dfrac{14}{9}\\ =\dfrac{-5}{9}+\dfrac{14}{9}\\ =1\)

AH
Akai Haruma
Giáo viên
1 tháng 3 2019

Lời giải:

a)

\(3x^2-5x+1=2x-3\)

\(\Leftrightarrow 3x^2-5x+1-2x+3=0\)

\(\Leftrightarrow 3x^2-7x+4=0\) (\(a=3; b=-7; c=4)\)

b)

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

\(\Leftrightarrow \frac{3}{5}x^2-4x-3-3x-\frac{1}{3}=0\)

\(\Leftrightarrow \frac{3}{5}x^2-7x-\frac{10}{3}=0(a=\frac{3}{5};b=-7; c=\frac{-10}{3})\)

c)

\(\Leftrightarrow -\sqrt{3}x^2+x-5-\sqrt{3}x-\sqrt{2}=0\)

\(\Leftrightarrow -\sqrt{3}x^2+(1-\sqrt{3})x-(5+\sqrt{2})=0\)

(\(a=-\sqrt{3}; b=1-\sqrt{3}; c=-(5+\sqrt{2}))\)

d)

\(\Leftrightarrow x^2-5(m+1)x+m^2-2=0\)

(\(a=1;b=-5(m+1); c=m^2-2)\)

 

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-(x+3/5)+4/5

=1/7-2x-3/5+4/5

=-2x+12/35

b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

-3/5<x<1/7

nên x-1/7<0; x+3/5>0

\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0

\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-(x+3/5)+4/5

=1/7-2x-3/5+4/5

=-2x+12/35

b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

-3/5<x<1/7

nên x-1/7<0; x+3/5>0

\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0

\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)

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