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\(\frac{x+46}{20}=x\frac{2}{5}\)
\(\Rightarrow\frac{x+46}{20}=\frac{5x+2}{5}=\frac{20x+8}{20}\)
=>x+46=20x+8
=>46-8=20x-x
=>38=19x
=>x=2
vậy x=2
a. 5x +2 4/5 = 15
=> 5x + 14/5 = 15
=> 5x = 15 - 14/5 = 61/5
=> x = 61/5 : 5
=> x = 61/25
a)5x + 2 4/5 =15
5x+14/5=15
5x=15- 14/5
5x=71/5
x=71/5 :5
x=71
a) \(\frac{-1}{5}\le\frac{x}{8}\le\frac{1}{4}\)
\(\frac{-8}{40}\le\frac{5x}{40}\le\frac{10}{40}\)
\(\Rightarrow-8\le5x\le10\)
\(\Rightarrow x\in\left\{-5;0;5;10\right\}\)
5.x = - 5
x = -5 ÷ 5 = -1
5.x = 0
=> x = 0
5.x = 5
=> x = 1
5.x = 10
=> x = 2
Vậy, \(x\in\left\{-1;0;1;2\right\}\)
Câu b) mình không biết làm, bạn thông cảm nha!!!
Cbht
a. \(-\frac{1}{5}\le\frac{x}{8}\le\frac{1}{4}\)
<=> \(-\frac{8}{40}\le\frac{5x}{40}\le\frac{10}{40}\)
<=> -8 \(\le\)5x \(\le\)10
<=> x + 5 \(\in\){\(\pm\)8;\(\pm\)7; \(\pm\)6;\(\pm\)5;\(\pm\)4;\(\pm\)3;\(\pm\)2; \(\pm\)1; 0; 9;10}
<=> x \(\in\left\{3;-13;2;-12;1;-11;0;-10;-1;-9;-2;-8;-3;-7;-4;-6;-5;4;5\right\}\)
b. \(\frac{x+46}{20}=x\frac{2}{5}\)
<=> 5x + 230 = 40x
<=> 35x = 230
<=> x = \(\frac{46}{7}\)
\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)
\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)
\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)
Vì 1/49+1/48+...+1/45 khác 0
Nên x+50=0
do đó x=-50
\(\left|a+2\right|=a\)
\(\Rightarrow a+2=\hept{\begin{cases}a\\-a\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a-a=2\\-a-a=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}0=2\left(loai\right)\\-2a=2\end{cases}}\)
\(\Rightarrow a=-1\)
\(\frac{a+4}{20}=\frac{5}{a+4}\)
\(\Rightarrow\left(a+4\right)^2=20\cdot5\)
\(\Rightarrow\left(a+4\right)^2=100\)
\(\Rightarrow\hept{\begin{cases}a+4=10\\a+4=-10\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=6\\a=-14\end{cases}}\)
\(x=2\)
tic cho mk nha
\(\frac{x+46}{20}=\frac{5x+2}{5}\)
=> (x+46).5=20.(5x+2)
=> 5x+230=100x+40
=> 5x-100x=40-230
=> -95x=-190
=> x=-190:(-95)
=> x=2