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a ) + ) Xét x \(\ge\) 3 => |x - 3| = x - 3
=> |x - 3| + x - 3 = x - 3 + x - 3 = 0
<=> 2x - 6 = 0
<=> 2x = 6
=> x = 3
+ ) Xét x < 3 => |x - 3| = 3 - x
=> |x - 3| + x - 3 = 3 - x + x - 3 = 0
<=> 0 = 0 ( loại )
Vậy x = 3
ý khác tương tự
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Tìm x:
b) 1/3.x+2/5.(x-1)=0
\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)
\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)
\(<=> x= \dfrac{6}{11}\)
Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases}
2x-3=0 \\
6-2x=0
\end{cases}\) \(<=> \begin{cases}
2x=3 \\
-2x=-6
\end{cases}\) \(<=>\begin{cases}
x=\dfrac{3}{2} \\
x=3
\end{cases}\)
Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2
\(<=> 2x-5= \dfrac{5}{2}\)
\(<=> 2x= \dfrac{15}{2}\)
\(<=> x= \dfrac{15}{4}\)
Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')
\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)
\(<=> \dfrac{1}{3}x =5\)
\(<=> x= 15\)
Vậy \(x= 15\)
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\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}=\frac{2.15}{5.11}=\frac{6}{11}\)
Vậy x = 6/11
a) \(\frac{1}{3}.x+\frac{2}{5}.\left(x-1\right)=0\)
\(\frac{1}{3}.x+\frac{2}{5}.x-\frac{2}{5}=0\)
\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)
\(x.\frac{11}{15}-\frac{2}{5}=0\)
\(x.\frac{11}{15}=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{11}{15}\)
\(x=\frac{6}{11}\)
b) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(3x-5x-\left(\frac{3}{2}+3\right)=x+\frac{1}{5}\)
\(-2x-\frac{9}{2}=x+\frac{1}{5}\)
\(\Rightarrow-2x-x=\frac{1}{5}+\frac{9}{2}\)
\(-3x=\frac{47}{10}\)
\(x=\frac{47}{10}:\left(-3\right)\)
\(x=\frac{-47}{30}\)
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Giải:
a) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(\Leftrightarrow\dfrac{11}{15}x-\dfrac{2}{5}=0\)
\(\Leftrightarrow\dfrac{11}{15}x=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{6}{11}\)
Vậy ...
b) \(3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=x+\dfrac{1}{5}\)
\(\Leftrightarrow3x-\dfrac{3}{2}-5x-3=x+\dfrac{1}{5}\)
\(\Leftrightarrow-2x-\dfrac{9}{2}=x+\dfrac{1}{5}\)
\(\Leftrightarrow-3x=\dfrac{47}{10}\)
\(\Leftrightarrow x=\dfrac{-47}{30}\)
Vậy ...
a, 1/3 . x + 2/5 . ( x - 1 ) = 0
1/3 . x + 2/5 . x - 2/5 = 0
x . ( 1/3 + 2/5 ) = 0 + 2/5
x . 11/15 = 2/5
x = 2/5 : 11/15
x = 6/11
b, 3 . ( x - 1/2 ) - 5 . ( x + 3/5 ) = x + 1/5
3 . x - 3 . 1/2 - 5 . x + 5. 3/5 = x + 1/5
3x - 3/2 - 5x + 3 = x + 1/5
3x - 5x + x = 1/5 + 3/2 - 3
-3x = -13/10
x = -13/10 : -1
x = -13/10
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a) \(x +(x + 1) + (x + 2) + ... + (x +30) = 620\)
\(=\left(x+x+...+x+x\right)+\left(1+2+...+30\right)\)
\(=31x+465=620\)
\(=31x=620-465\)
\(=31x=155\)
\(=x=155\div31\)
\(x=5\)
b) \(2+4+6+8+....+2x = 210\)
\(\Rightarrow2.1+2.2+2.3+2.4+...+2.x\)
\(\Rightarrow2.\left(2+4+6+8+...+x\right)=210\)
\(\Rightarrow2+4+6+8+x=210\div2\)
\(\Rightarrow2+4+6+8+...+x=105\)
\(\Rightarrow x=14\)
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a) \(\left(x-2\right).\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)
b) \(\left(3x+9\right).\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-9\\3x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
c) (31 - 2x)3 =27
(31 - 2x)3 = 33
=> 31 - 2x = 3
2x = 31 - 3
2x = 28
x = 14
a. \(\left(x-2\right).\left(2x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=2\)hoặc \(x=\frac{1}{2}\)
b.\(\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}}\)
Vậy \(x=-3\)hoặc \(x=\frac{1}{3}\)
c.\(\left(31-2x\right)^3=-27\)
\(\Leftrightarrow\left(31-2x\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow31-2x=-3\)
\(2x=34\)
\(x=17\)
d.\(\left(x-2\right).\left(7-x\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}}\)
Vậy \(x=2\)hoặc \(x=7\)
e.\(\left(x-5\right)^5=32\)
\(\Leftrightarrow\left(x-5\right)^5=2^5\)
\(\Leftrightarrow x-5=2\Leftrightarrow x=7\)
f.\(\left(2-x\right)^4=81\)
\(\Leftrightarrow\left(2-x\right)^4=3^4\)
\(2-x=3\Leftrightarrow x=-1\)
g.\(\left|x-7\right|< 3\Leftrightarrow-3< x-7< 3\Leftrightarrow4< x< 10\)
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e) \(\left(x-3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left(x-3\right)=0\) ( \(x^2+1>0\forall x\))
\(\Rightarrow x=3\)
đ) \(4.8^2=2^x\)
\(2^2.\left(2^3\right)^2=2^x\)
\(2^2.2^6=2^x\)
\(2^8=2^x\)
\(\Rightarrow x=8\)
d) \(\left|x+3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)
mấy câu trên dễ rồi tự làm em nhé
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a, 45-15=-75-x
=> 30 = -75 - x
=> x = -75 - 30
=> x = -105
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\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
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Bài 1 : \(a,\left|x-3,5\right|=7,5\)
\(\Rightarrow\orbr{\begin{cases}x-3,5=7,5\\x-3,5=-7,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-4\end{cases}}\)
\(b,\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
\(c,3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=3,6\\x-0,4=-3,6\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3,2\end{cases}}\)
\(d,\left|x-\frac{1}{2}\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{4}{3}\\x-\frac{1}{2}=-\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{6}\\x=-\frac{5}{6}\end{cases}}\)
\(\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{x}{3}\right)=0\)
=>\(\left(x+\dfrac{1}{2}\right)\cdot\dfrac{2x}{3}=0\)
=>\(x\left(x+\dfrac{1}{2}\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)