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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
\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)
\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)
\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)
Vì 1/49+1/48+...+1/45 khác 0
Nên x+50=0
do đó x=-50
Ta có :
\(\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-4}{46}+\frac{x-5}{45}=5\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{49}-1\right)+\left(\frac{x-2}{48}-1\right)+\left(\frac{x-3}{47}-1\right)+\left(\frac{x-4}{46}-1\right)+\left(\frac{x-5}{45}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-1-49}{49}+\frac{x-2-48}{48}+\frac{x-3-47}{47}+\frac{x-4-46}{46}+\frac{x-5-45}{45}=0\)
\(\Leftrightarrow\)\(\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{46}+\frac{x-50}{45}=0\)
\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\) ( vì nó lớn hơn 0 )
Nên \(x-50=0\)
\(\Rightarrow\)\(x=50\)
Vậy \(x=50\)
Chúc bạn học tốt ~
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 = \(\frac{46}{7}\)
\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)
\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)
ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP