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a, x^2 - 2x + 7
= x( x-2) + 7
ta có x(x-2) chia hết cho x- 2
nên để x^2 - 2x + 7 chia hết cho 2
thì 7 chia hết cho x- 2
=> x-2 thuộc ước của 7
đến đây tự làm tiếp
a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)
=> \(\frac{2}{3}:x=-7-\frac{1}{3}\)
=> \(\frac{2}{3}:x=-\frac{22}{3}\)
=> \(x=\frac{2}{3}:\left(-\frac{22}{3}\right)\)
=> \(x=-\frac{1}{11}\)
b) \(\frac{1}{3}x+\frac{2}{5}x=0\)
=> \(\frac{11}{15}x=0\)
=> \(x=0\)
c) \(\left(2x-3\right)\left(6-2x\right)=0\)
=> \(\left(2x-3\right)\left(3-x\right).2=0\)
=> \(\left(2x-3\right)\left(3-x\right)=0\)
=> \(\orbr{\begin{cases}2x-3=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)
\(\Rightarrow\frac{2}{3}.\frac{1}{x}=-7-\frac{1}{3}\)
\(\Rightarrow\frac{2}{3x}=\frac{-21-1}{3}\)
\(\Rightarrow\frac{2}{3x}=\frac{-22}{3}\)
\(\Rightarrow-22.3x=6\)
\(\Rightarrow3x=\frac{-6}{22}=\frac{-3}{11}\)
\(\Rightarrow x=\frac{-3}{11}:3=\frac{-3}{11}.\frac{1}{3}\)
\(\Rightarrow x=\frac{-1}{11}\)
b) \(\frac{1}{3}x+\frac{2}{5}x=0\)
\(\Rightarrow x.\left(\frac{1}{3}+\frac{2}{5}\right)=0\)
\(\Rightarrow x=0\)
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=3\\2x=6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
d) \(x:\frac{3}{4}+\frac{1}{4}=\frac{-2}{3}\)
\(\Rightarrow x.\frac{4}{3}=\frac{-2}{3}-\frac{1}{4}\)
\(\Rightarrow x.\frac{4}{3}=\frac{-11}{12}\)
\(\Rightarrow x=\frac{-11}{12}:\frac{4}{3}=\frac{-11}{12}.\frac{3}{4}=\frac{-11}{16}\)
e) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{4}\\x-\frac{2}{3}=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)
\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)
\(\frac{62}{7}:x=\frac{1624}{27}\)
\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)
\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)
\(\frac{1}{5}:x=\frac{2}{35}\)
\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)
\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)
\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)
\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)
\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)
\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)
\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)
\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)
\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)
\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)
\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)
\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)
\(1-x=\frac{2}{5}:\frac{2}{5}=1\)
\(x=1-1=0\)
\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)
\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)
\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)
học tốt nha
\(\left|x-1\right|+3.\left|x-3\right|-2.\left|x-2\right|=4\)
Ta thấy \(\left|x-1\right|\ge0;\left|x-3\right|\ge0;\left|x-2\right|\ge0\Rightarrow\left|x-1\right|\ge0;3.\left|x-3\right|\ge0;2.\left|x-2\right|\ge0\)
Khi đó : \(\left|x-1\right|+3.\left|x-3\right|-2.\left|x-2\right|=4\)
\(\Rightarrow x-1+3.\left(x-3\right)-2.\left(x-2\right)=4\)
\(\Rightarrow x-1+3x-3-4x-2=4\)
Tự giải tiếp nhé , tìm x bình thường
a, (x + 2) + (x + 4) + (x + 6) + ... + (x + 50) = 750
=> x + 2 + x + 4 + x + 6 + ... + x + 50 = 750
=> (x + x + x + ... + x) + (2 + 4 + 6 + ... + 50) = 750
=> 25x + (50 + 2).25 : 2 = 750
=> 25x + 52.25 : 2 = 750
=> 25x + 650 = 750
=> 25x = 100
=> x = 4
a) ( x+x+...+x)+(2+4+6+...+50)= 750
( x*25)+ (50+2)*25:2 = 750
(x*25)+ 650 = 750
x* 25 = 750 - 650 = 100
x = 100 :25 = 4
\(3|x-\frac{1}{2}|+\frac{3}{4}=-2|x-\frac{1}{2}|\)
\(\Rightarrow\) \(3|x-\frac{1}{2}|+2|x-\frac{1}{2}|=-\frac{3}{4}\)
\(\Rightarrow5|x-\frac{1}{2}|=-\frac{3}{4}\)
\(\Rightarrow|x-\frac{1}{2}|=-\frac{3}{4}:5=-\frac{3}{20}\) ( vô lý )
Vậy ko tồn tại x thỏa mãn yêu cầu bài toán
\(3.\left|x-\frac{1}{2}\right|+\frac{3}{4}=-2.\left|\frac{1}{2}-x\right|\)
\(x-\frac{1}{2}\ge0\) cho: \(x\ge\frac{1}{2}\) do đó: \(x\ge\frac{1}{2};\left|x-\frac{1}{2}\right|=x-\frac{1}{2}\)
\(x-\frac{1}{2}< 0\) cho: \(x< \frac{1}{2}\) do đó: \(x\le\frac{1}{2};\left|x-\frac{1}{2}\right|=-\left(x-\frac{1}{2}\right)\)
\(\frac{1}{2}-x\ge0\) cho \(x\le\frac{1}{2}\) do đó: \(x\le\frac{1}{2};\left|\frac{1}{2}-x\right|=\frac{1}{2}-x\)
\(\frac{1}{2}-x< 0\) cho \(x>\frac{1}{2}\) do đó: \(x>\frac{1}{2}\left|\frac{1}{2}-x\right|=-\left(\frac{1}{2}-x\right)\)
\(x< \frac{1}{2};x\ge\frac{1}{2}\)
Ta xét 2th:
Th1: \(3\left[-\left(x-\frac{1}{2}\right)\right]+\frac{3}{4}=-2\left(\frac{1}{2}-x\right)\)
\(x=\frac{13}{20}\) (loại)
Th2: \(3\left(x-\frac{1}{2}\right)+\frac{3}{4}=2\left[-\left(\frac{1}{2}-x\right)\right]\)
\(x=\frac{7}{20}\) (loại)
=> Không có giá trị thỏa mãn đề bài.