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tìm x , biết :
a . 1,5 . x + 2,5 = 1
\(1,5.x=-1,5\)
\(x=\dfrac{-1,5}{1,5}\)
\(x=-1\)
b . ( 0,5 . x - 1,3 ) : 0,5 = -1,7
\(0,5.x-1,3=-0,85\)
\(0,5x=-2,15\)
\(x=\dfrac{-2,15}{0,5}\)
\(x=-4,3\)
c . 1,75 .x - 1,5 . x = 0,84
\(x\left(1,75-1,5\right)=0,84\)
\(x.0,25=0,84\)
\(x=3,36\)
d . x + 75% . x + 125% = 1
\(x+\dfrac{75}{100}.x+\dfrac{125}{100}=1\)
\(x+\dfrac{3}{4}.x+\dfrac{5}{4}=1\)
\(x+\dfrac{3}{4}.x=-\dfrac{1}{4}\)
\(x.\left(1+\dfrac{3}{4}\right)=-\dfrac{1}{4}\)
\(x.1,75=-\dfrac{1}{4}\)
\(x=-\dfrac{1}{7}\)
e . x - 25% .x + 75% = 1
\(x-\dfrac{25}{100}.x+\dfrac{75}{100}=1\)
\(x-\dfrac{1}{4}.x+\dfrac{3}{4}=1\)
\(x-\dfrac{1}{4}.x=\dfrac{1}{4}\)
\(x.\left(1-\dfrac{1}{4}\right)=\dfrac{1}{4}\)
\(x.\dfrac{3}{4}=\dfrac{1}{4}\)
\(x=\dfrac{1}{3}\)
f . | x - 1,5 | = 2,37
TH1 : \(x-1,5=2,37\)
\(x=2,37+1,5\)
\(x=3,87\)
TH2 : \(x-1,5=-2,37\)
\(x=-2,37+1,5\)
\(x=-0,87\)
a) \(\left|3,5-x\right|=1,3\)
\(\Rightarrow\left[{}\begin{matrix}3,5-x=1,3\\3,5-x=-1,3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3,5-1,3\\x=3,5+1,3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2,2\\x=4,8\end{matrix}\right.\)
b) \(1,6-\left|x-0,2\right|=0,4\)
\(\Rightarrow\left|x-0,2\right|=1,2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,2\\x-0,2=-1,2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1,2+0,2\\x=-1,2+0,2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1,4\\x=-1\end{matrix}\right.\)
\(\left|3,5-x\right|=1,3\)
\(\Rightarrow\left[{}\begin{matrix}3,5-x=1,3\Rightarrow x=2,2\\3,5-x=-1,3\Rightarrow x=4,8\end{matrix}\right.\)
\(1,6-\left|x-0,2\right|=0,4\)
\(\Rightarrow\left|x-0,2\right|=1,2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,2\Rightarrow x=1,4\\x-0,2=-1,2\Rightarrow x=-1\end{matrix}\right.\)
\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-1,5\right|=0\Rightarrow x=1,5\\\left|2,5-x\right|=0\Rightarrow x=2,5\end{matrix}\right.\)
\(1,5\ne2,5\Rightarrow x\in\varnothing\)
a) TH1: \(x< 2,5\) , ta có:
\(2,5-x=1,3\)
\(x=2,5-1,3=1,2\)
TH2: \(x\ge2,5\), ta có:
\(x-2,5=1,3\)
\(x=1,3+2,5=3,8\)
Vậy \(\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)
b) \(1,6-\left|x-0,2\right|=0\)
\(\Rightarrow\left|x-0,2\right|=1,6\)
TH1: \(x< 0,2\) ta có:
\(0,2-x=1,6\Rightarrow x=0,2-1,6=-1,4\)
TH2: \(x\ge0,2\) ta có:
\(x-0,2=1,6\Rightarrow x=1,6+0,2=1,8\)
Vậy \(\orbr{\begin{cases}x=-1,4\\x=1,8\end{cases}}\)
(\(x\) - 2,5) \(\times\) 8 = \(x\) \(\times\) 6 - 4
8\(x\) - 20 = 6\(x\) - 4
8\(x\) - 6\(x\) = 20 - 4
2\(x\) = 16
\(x\) = 16 : 2
\(x\) = 8
\(a,x+3,79=15,12\)\(\Rightarrow x=15,12-3,79\)\(\Rightarrow x=11,33\)
\(b,12,1-x=5,49\)\(\Rightarrow x=12,1-5,49\Rightarrow x=6,61\)
\(c,60,45:x=1,5\)\(\Rightarrow x=60,45:1,5\Rightarrow x=40,3\)
\(d,3,5.x-1,5.x=12,5\)
\(\Rightarrow x\left(3,5-1,5\right)=12,5\)
\(\Rightarrow x.2=12,25\Rightarrow x=12,25:2=6,125\)
\(d,12,5.x-2,5.x-50=2,5.10\)
\(\Rightarrow12,5.x-2,5.x-50=25\)
\(\Rightarrow x\left(12,5-2,5\right)=25+50\)
\(\Rightarrow x.10=75\)
\(\Rightarrow x=75:10=7,5\)
a) x=11,33 b) x=6,61 c)40,3
Mình chỉ biết làm tới đó thôi. xin lỗi nha
a: =>|x-1/2|=2x+1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(2x+1\right)^2-\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(2x+1-x+\dfrac{1}{2}\right)\left(2x+1+x-\dfrac{1}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(x+\dfrac{3}{2}\right)\left(3x-\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
b: =>\(\left\{{}\begin{matrix}x-1.3=0\\2y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1.3\\y=\dfrac{1}{2}\end{matrix}\right.\)
a . ( x - 1/2 ) - 2 x = 1
=> x - 1/2 = 1 hoặc 2x =0
=> x = 3/2 hoặc x = 0
b .( x -1/3 ) + ( 2y -1 ) = 0
=> x - 1/3 = 0 hoặc 2y - 1 = 0
=> x = 1/3 hoặc 2y = 1
=> x = 1/3 hoặc y = 1/2
c. ( x - 1,5 ) + ( y - 2,5 ) + ( x + y + z ) nhỏ hơn hoặc bằng 0
=> x - 1,5 = 0 hoặc y - 2,5 = 0 hoặc x + y + z = 0
=> x= 1,5 hoặc y= 2,5 hoặc x + y +z = 0
=> x = 1,5 hoặc y = 2,5 hoặc 1,5 + 2,5 + z = 0
=> x = 1,5 hoặc y = 2,5 hoặc z = 4 , - 4
2 cái đầu tính bình thường nha lớp 6 rồi kg tính đc thì ...cx cạn lời lun
còn 2 cái đằng sau thì suy ra cộng trừ rồi giải 2 trường hợp