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\(\left|3-x\right|=x-5\)
\(\Rightarrow\orbr{\begin{cases}3-x=x-5\\3-x=5-x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x-x=-5-3\\-x+x=5-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=-8\\x\in\varnothing\end{cases}}\)
\(\Rightarrow x=4\)
vậy_
1) \(\left|3-x\right|=x-5\)
\(3x-x\ge0\text{ để: }x\ge0\Rightarrow x\ge0;\left|3x-x\right|=3x-x\)
\(3x-x< 0\text{ để: }x< 0\Rightarrow\left|3x-x\right|=-\left(3x-x\right)\)
\(\Rightarrow\orbr{\begin{cases}x< 0\\x\ge0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-5\end{cases}}\)
=> Không có gtrị tmyk.
Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\)
=> \(A=\frac{9}{10}\)
b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> \(A=1+\frac{7}{n-5}\)
Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)
=> n=(-2; 4, 6, 8)
\(a,\left[\frac{4}{5}+\frac{2}{3}\right]:\frac{1}{5}-1,4\cdot\left[\frac{-5}{7}\right]^2\)
\(=\left[\frac{4\cdot3}{15}+\frac{2\cdot5}{15}\right]:\frac{1}{5}-1,4\cdot\frac{-5}{7}\cdot\frac{-5}{7}\)
\(=\left[\frac{12}{15}+\frac{10}{15}\right]:\frac{1}{5}-\frac{14}{10}\cdot\frac{25}{49}\)
\(=\frac{22}{15}:\frac{1}{5}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22}{15}\cdot\frac{5}{1}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22\cdot5}{15\cdot1}-\frac{7\cdot25}{5\cdot49}=\frac{22\cdot1}{3\cdot1}-\frac{1\cdot5}{1\cdot7}=\frac{22}{3}-\frac{5}{7}\)
= ...
Tự tính
Bài 2 : \(a,3-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}+3\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{2}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{23}{3}\\x=\frac{-19}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{23}{3};\frac{-19}{3}\right\}\)
b, \(0,6-160\%< x\le3\frac{2}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{160}{100}< x\le\frac{11}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{8}{5}< x\le\frac{11}{3}\cdot\frac{18}{22}\)
\(\Rightarrow0,6-1,6< x\le3\)
\(\Rightarrow-1< x\le3\)
\(\Rightarrow x\in\left\{0;1;2;3\right\}\)
\(\frac{x-2}{4}=\frac{-9}{2-x}\)
\(\Rightarrow\frac{x-2}{4}=\frac{9}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=\left(x+2\right)5\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
c;giống câu trên :v
câu 1b
Gọi d là ƯCLN (3n-7, 2n-5), d thuộc N*
Ta có : 3n-7 chia ht cho d , 2n_5 chia ht cho d
suy ra: 2(3n-7) chia ht cho d , 3(2n-5) chia ht cho d
suy ra 6n-14 chia ht cho d, 6n-15 chia ht cho d
dấu suy ra [(6n -15) - (6n-14)] chia ht cho d dấu suy ra 1 chia ht cho d suy ra d =1
Vậy......
1) b. Để chứng tỏ \(\frac{3n-7}{2n-5}\) là phân số tối giản
Ta cần chứng minh: ( 3n - 7; 2n - 5 ) = 1
Thật vậy: ( 3n - 7 ; 2n - 5 ) = ( 2n - 5 ; ( 3n - 7 ) - ( 2n - 5 ) ) = ( 2n - 5; n - 2 ) = ( n - 2; n - 3 ) = ( n - 2; 1 ) = 1
=> \(\frac{3n-7}{2n-5}\) là phân số tối giản
3) \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{12}\)
Ta có: \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\left(\frac{1}{5}+\frac{1}{7}\right)+\frac{1}{6}=\frac{12}{35}+\frac{1}{6}>\frac{12}{36}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{1}{2}\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}=\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}\right)>\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \)
=> A > 1/2 + 1/2 + 1/2 + 1/2 = 2
\(\frac{5}{2}x+\frac{1}{2}x=x+400\%\)
\(\Rightarrow\left(\frac{5}{2}+\frac{1}{2}\right)x=x+4\)
\(\Rightarrow\frac{6}{2}x=x+4\)
\(\Rightarrow3x=x+4\)
\(\Rightarrow3x-x=4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Chúc bạn học tốt !!!
<=> \(\frac{x-1}{2}=\frac{3-5}{x-1}\)
<=> \(\left(x-1\right)^2=-4\)
Vì \(VT\ge0\) Mà \(VP< 0\)
=> Pt vô nghiệm.