\(\hept{\begin{cases}64x+56y=29,6\\3x+4y=1,8\end{cases}}\) 

\(\hept{\begin{cases}64x+56y=29,6\\-42x-56y=-25,2\end{cases}}\)   

\(\hept{\begin{cases}22x=4,4\\64x+56y=29,6\end{cases}}\) 

\(\hept{\begin{cases}x=0,2\\y=0,3\end{cases}}\)

\(\hept{\begin{cases}64x+56y=29,6\left(1\right)\\\frac{x}{2}+\frac{2y}{3}=0,3\left(2\right)\end{cases}}\)

Nhân 128 vào từng vế của 2

hpt <=> \(\hept{\begin{cases}64x+56y=29,6\\64x+\frac{256}{3}y=38,4\left(3\right)\end{cases}}\)

Lấy (1) trừ (3) theo vế

\(\Leftrightarrow-\frac{88}{3}y=-\frac{44}{5}\Leftrightarrow y=\frac{3}{10}\)

Thế y = 3/10 vào (1)

\(\Leftrightarrow64x+56\cdot\frac{3}{10}=29,6\)

\(\Leftrightarrow64x+\frac{84}{5}=29,6\)

\(\Leftrightarrow64x=\frac{64}{5}\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2 ; y = 3/10

Hơi dài dòng xíu :>

x = 0,2 ; y = 0,3

\(1)\) \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Leftrightarrow\)\(3^x.1+3^x.3+3^x.3^2=351\)

\(\Leftrightarrow\)\(3^x\left(1+3+3^2\right)=351\)

\(\Leftrightarrow\)\(3^x.13=351\)

\(\Leftrightarrow\)\(3^x=\frac{351}{13}\)

\(\Leftrightarrow\)\(3^x=27\)

\(\Leftrightarrow\)\(3^x=3^3\)

\(\Leftrightarrow\)\(x=3\)

Vậy \(x=3\)

Chúc bạn học tốt ~ 

\(2)\) 

\(a)\) Ta có : 

\(25^{15}=\left(5^2\right)^{15}=5^{2.15}=5^{30}\)

\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}=2^{30}.3^{30}=\left(2.3\right)^{30}=6^{30}\)

Vì \(5^{30}< 6^{30}\) nên \(25^{15}< 8^{10}.3^{30}\)

Vậy \(25^{15}< 8^{10}.3^{30}\)

\(b)\) Ta có : 

\(\left(0,3\right)^{20}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)

Vì \(\left(0,1\right)^{10}>\left(0,09\right)^{10}\) nên \(\left(0,1\right)^{10}>\left(0,3\right)^{20}\)

Vậy \(\left(0,1\right)^{10}>\left(0,3\right)^{20}\)

Chúc bạn học tốt ~ 

x=-1               

x=-1                                 

x = -1

ai tk mk

mk tk lại

hứa luôn

thank nhiều

a)2,3:x+3,4:x=6

=>5,7/x=6

=>x=0,95

b)10,2:x+4,5:x=14

=>14,7:x=14

=>x=1,05

c)Xx4,9+x:10

=>49xX:10+x:10=102:10

=>50xX=102

=>x=2,04

d)4,1x-x:10=2,4

=>41x:10-x:10=24:10

=>40x=24

=>x=0,6

còn lại bạn tự làm nha mỏi tay quá

nhớ
 

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=0\)

\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)

\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

\(\Rightarrow x-100=0\left(Vì\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\right)\)

\(\Rightarrow x=100\)

may gioi vai

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)

\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\)\(\ne\)0 nên x + 1 = 0 \(\Rightarrow\)x = -1

-1

​giúp con với ông bà các bác ơi

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(\Rightarrow x.0+1=0\)

=> 1=0 ( Vô lý )

Vậy \(x\in\varnothing\)

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow x.\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)=1\)

\(\Rightarrow x.0=1\Rightarrow x=0\)

Vậy x=0