nhanh nha các bn

Ta có :

a) \(\frac{21^{30}}{63^{15}}=\frac{\left(21^2\right)^{15}}{63^{15}}=\left(\frac{21^2}{63}\right)^{15}\)

b) \(\frac{28^9.30^9}{105^9}=\frac{\left(28.30\right)^9}{105^9}=\frac{840^9}{105^9}=\left(\frac{840}{105}\right)^9=8^9\)

6,3 + (-3,7) + 2,4 + (-0,3)

= (6,3 + 2,4) + [ ( - 3,7 ) + ( - 0,3) ] 

= 8,7 +  [ - ( 3,7 + 0,3 ) ]

= 8,7 + ( - 4 )

= 4,7

a) 6,3 + ( - 3,7 ) + 2,4 + ( - 0,3 )

= 6,3 - 0,3 + 2,4 - 3,7 = 6 + 2,4 - 3,7

= 8,4 - 3,7 = 4,7

a.-5,639

b.-0,32

c.16,027

d.-2,16

sorry làm lại nè

\(\left(\frac{1}{2}\right)^4và\left(\frac{1}{4}\right)^4\)

ta thấy : \(\frac{1}{2}>\frac{1}{4}\Rightarrow\left(\frac{1}{2}\right)^4>\left(\frac{1}{4}\right)^4\)

Nhanh nha mọi người

\(\text{|}5-\frac{3}{4}x\text{|}+\text{|}\frac{2}{7}y-3\text{|}=0\)

Mà \(\text{|}5-\frac{3}{4}x\text{|}\ge0\)và \(\text{|}\frac{2}{7}y-3\text{|}\ge0\)

=>\(\hept{\begin{cases}\text{|}5-\frac{3}{4}x\text{|}=0\\\text{|}\frac{2}{7}y-3\text{|}=0\end{cases}}\)

<=>\(\hept{\begin{cases}\text{ }5-\frac{3}{4}x\text{ }=0\\\text{ }\frac{2}{7}y-3\text{ }=0\end{cases}}\)

<=>\(\hept{\begin{cases}\text{ }\frac{3}{4}x\text{ }=5\\\text{ }\frac{2}{7}y\text{ }=3\end{cases}}\)

<=>\(\hept{\begin{cases}\text{ }x\text{ }=5:\frac{3}{4}=\frac{20}{3}\\\text{ }y\text{ }=3:\frac{2}{7}=\frac{21}{2}\end{cases}}\)

\(\left|5-\frac{3}{4}x\right|+\left|\frac{2}{7}y-3\right|=0\)

\(\Rightarrow\hept{\begin{cases}5-\frac{3}{4}x=0\\\frac{2}{7}y-3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{20}{3}\\y=\frac{21}{2}\end{cases}}\)

$\left|3-2x\right|+\left|4y+5\right|=0$

$\Rightarrow \left|3-2x\right|=0$ và $\left|4y+5\right|=0$

+) $\left|3-2x\right|=0\Rightarrow 3-2x=0\Rightarrow x=\frac{3}{2}$

+) $\left|4y+5\right|=0\Rightarrow 4y+5=0\Rightarrow y=\frac{5}{4}$

Vậy $x=\frac{3}{2};y=\frac{5}{4}$

\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}-\frac{3}{4}=\text{|}-\frac{3}{4}\text{|}\)

<=>\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}-\frac{3}{4}=\frac{3}{4}\text{ }\)

<=>\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}=\frac{3}{4}+\frac{3}{4}\text{ }\)

<=>\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}=\frac{3}{2}\)

<=>\(\orbr{\begin{cases}\text{ }\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\\frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{cases}}\)

<=>\(\orbr{\begin{cases}\text{ }\frac{3}{4}x=\frac{3}{2}+\frac{3}{4}\\\frac{3}{4}x=-\frac{3}{2}+\frac{3}{4}\end{cases}}\)

<=>\(\orbr{\begin{cases}\text{ }\frac{3}{4}x=\frac{9}{4}\\\frac{3}{4}x=-\frac{3}{4}\end{cases}}\)

<=>\(\orbr{\begin{cases}\text{ }x=\frac{9}{4}:\frac{3}{4}=3\\x=-\frac{3}{4}:\frac{3}{4}=-1\end{cases}}\)

\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|\frac{-3}{4}\right|\)

\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)

\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)

\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x-3=\frac{3}{2}\\\frac{3}{4}x-3=\frac{-3}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x=\frac{9}{4}\\\frac{3}{4}x=\frac{-3}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Ta có:

| x + 3/5 | - | x - 7/3 | = 0

=> | x + 3/5 |=| x - 7/3 |

=>x+3/5=x-7/3 hoặc x+3/5 = -(x-7/3)

=>x-x=-7/3-3/5 hoặc x+3/5 =-x+7/3

=>0=-7/3-3/5 hoặc x+x=7/3-3/5

=>(vô lí) hoặc 2x=24/15=8/5

=>x=8/5.1/2

=>x=4/5.

Vậy....

Ta có :

\(\left|x+\frac{3}{5}\right|-\left|x-\frac{7}{3}\right|=0\)

\(\Rightarrow\left|x+\frac{3}{5}\right|=\left|x-\frac{7}{3}\right|\)

Ta xét 3 trường hợp :

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

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

\(\Rightarrow\frac{3}{5}+\frac{7}{3}=0\left(\text{vô lý}\right)\)

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

\(\Rightarrow x+x=\frac{7}{3}+\frac{3}{5}\)

\(\Rightarrow2.x=2\frac{14}{15}\)

\(\Rightarrow x=2\frac{14}{15}:2=1\frac{7}{15}\)

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

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

\(\Rightarrow1\frac{11}{15}=2.x\)

\(\Rightarrow x=1\frac{11}{15}:2=\frac{13}{15}\)

Ta có: | x + 3 | = 3x - 1

+ Với x ≥ - 3, phương trình tương đương: x + 3 = 3x + 1 ⇔ - 2x = - 2 ⇔ x = 1.

Thỏa mãn điều kiện x ≥ - 3

+ Với x < - 3, phương trình tương đương: - x - 3 = 3x + 1 ⇔ - 4x = 4 ⇔ x = - 1

Không thỏa mã điều kiện x < - 3

Vậy phương trình đã cho có tập nghiệm là S = { 1 }