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=>|x-3|+5|x-3|=360
=>|x-3|=60
=>x-3=60 hoặc x-3=-60
=>x=63 hoặc x=-57
b ) 2. | x | - 15= -7
2. | x | = -7 + 15
2. | x | = 8
| x | = 8 : 2
| x | = 4
⇒ x = 4 , x = -4
Vậy x = 4 hoặc - 4
c) ( - 187 ) + 1948- ( -287 ) + | -1948 | + | - 1918 |
= ( - 187 ) + 1948- ( -287 ) + 1948 + 1918
= 1761- 1211+ 3866
= 550 + 3866
= 4416
\(\dfrac{\left(-9\right).11+32.\left(-9\right)}{\left(-43\right)15+12\left(-43\right)}\\ =\dfrac{\left(-9\right)\left(11+32\right)}{\left(-43\right)\left(15+12\right)}\\ =\dfrac{9\left(-43\right)}{\left(-43\right)27}\\ =\dfrac{9}{27}\\ =\dfrac{1}{3}\)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Rightarrow2x-1=0\)hoặc \(2x-1=1\)hoặc \(2x-1=-1\)
\(\Rightarrow x=\frac{1}{2}\)hoặc \(x=1\)hoặc \(x=0\)
a) \(56+\left(-29\right)+\left(-7\right)+28+13+\left(-35\right)\)
\(=27+\left(-7\right)+28+13+\left(-35\right)\)
\(=20+28+13+\left(-35\right)\)
\(=48+13+\left(-35\right)\)
\(=61+\left(-35\right)\)
\(=26\)
b) \(\left(-213\right)+186+\left(-14\right)+217+54+\left(-49\right)\)
\(=\left(-27\right)+\left(-14\right)+217+54+\left(-49\right)\)
\(=\left(-41\right)+217+54+\left(-49\right)\)
\(=176+54+\left(-49\right)\)
\(=230+\left(-49\right)\)
\(=181\)
c) \(435+\left(-43\right)+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
\(=392+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
\(=\left(-91\right)+\left(-57\right)+383+\left(-415\right)\)
\(=\left(-148\right)+383+\left(-415\right)\)
\(=235+\left(-415\right)\)
\(=-180\)
Lời giải:
a)
\(-3\frac{5}{8}+\left(-\frac{3}{8}+\frac{9}{4}\right)\)
\(=-\frac{29}{8}+\left(-\frac{3}{8}+\frac{18}{8}\right)\)
\(=-\frac{29}{8}+\frac{15}{8}=-\frac{14}{8}=-\frac{7}{4}\)
b) \(\frac{\left(-9\right)\cdot11+32\cdot\left(-9\right)}{\left(-43\right)\cdot15+12\cdot\left(-43\right)}=\frac{\left(-9\right)\left(11+32\right)}{\left(-43\right)\left(15+12\right)}=\frac{\left(-9\right)\cdot43}{\left(-43\right)\cdot27}=\frac{\left(-1\right)\cdot1}{\left(-1\right)\cdot3}=\frac{1}{3}\)
c) Thay \(x=\frac{2011}{2012}\)vào biểu thức \(x\cdot\frac{1}{3}+2x\cdot\frac{3}{6}-3x\cdot\frac{4}{9}\)ta có :
\(\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{3}{6}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)
\(=\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{1}{2}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)
\(=\frac{2011}{6036}+\frac{2011}{2012}-\frac{2011}{1509}\)
\(=\frac{2011}{6036}+\frac{6033}{6036}-\frac{8044}{6036}=\frac{2011+6033-8044}{6036}=0\)
1/Ta co :
\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}\right)+\dfrac{46}{45}.\left(\dfrac{15}{7}-\dfrac{45}{46}\right)\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-\dfrac{15}{7}.\dfrac{46}{45}+\dfrac{46}{45}.\dfrac{15}{7}-\dfrac{46}{45}.\dfrac{45}{46}\)
=\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}+\dfrac{46}{45}\right)-1\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-1\)
=\(\dfrac{3}{7}-1=\dfrac{-4}{7}\)
2/Ta co
\(\dfrac{43}{47}.\left(\dfrac{18}{37}+\dfrac{47}{43}\right)-\dfrac{18}{3}.\left(\dfrac{43}{47}+\dfrac{37}{36}\right)\)
=\(\dfrac{43}{47}.\dfrac{18}{37}+\dfrac{43}{47}.\dfrac{47}{43}-\dfrac{18}{37}.\dfrac{43}{47}+\dfrac{18}{37}.\dfrac{37}{36}\)
=\(\dfrac{18}{37}.\left(\dfrac{43}{37}-\dfrac{43}{37}+\dfrac{37}{36}\right)+1\)
=\(\dfrac{18}{37}.\dfrac{37}{36}+1\)
=\(\dfrac{1}{2}+1=\dfrac{3}{2}\)
tick cho mk nha
giúp mình đi
345 + (287 - 78 + 43) - (345 + 287 - 43)
= 345 + 287 - 78 + 43 - 345 - 287 + 43
= (345 - 345) + (287 - 287) + (43 + 43 - 78)
= 0 + 0 + (86 - 78)
= 8