1 tìm x
a x.3+7=16
b x-152:2+46
c 74-2.(x+3)=34
giúp mik vs ạ
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
ND
4
24 tháng 8 2023
\(x-\dfrac{1}{3}=\dfrac{2}{3}.\dfrac{9}{14}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{4}{7}\)
\(x=\dfrac{19}{21}\)
KL
2
22 tháng 10 2023
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
NM
Nguyễn Minh Quang
Giáo viên
14 tháng 8 2021
ta có :
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2+z^2}{4+9+25}=\frac{152}{38}=4\)
vậy ta có \(x^2=16\Rightarrow\orbr{\begin{cases}x=4,y=-6,z=10\\x=-4,y=6,z=-10\end{cases}}\)
14 tháng 10 2021
a) \(\Rightarrow5\left(x-10\right)=10\)
\(\Rightarrow x-10=2\Rightarrow x=12\)
b) \(\Rightarrow3\left(70-x\right)+5=92\)
\(\Rightarrow3\left(70-x\right)=87\)
\(\Rightarrow70-x=29\Rightarrow x=41\)
c) \(\Rightarrow230+\left[16+\left(x-5\right)\right]=315\)
\(\Rightarrow11+x=85\Rightarrow x=74\)
d) \(\Rightarrow707:\left(2^x-5+74\right)=7\)
\(\Rightarrow2^x-5+74=101\Rightarrow2^x-5=27\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
NH
15 tháng 12 2022
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
DT
2
7 tháng 3 2022
(2x - 1)^2 + (x + 3)^2 - 5(x + 7)(x - 7) = 0
<=>4x^2-4x+1+x^2+6x+9-5x^2+245=0
<=>2x+255=0
<=>2x=-255
<=>x=-255/2
LH
4
NM
Nguyễn Minh Quang
Giáo viên
4 tháng 8 2021
ta có \(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)
\(\Leftrightarrow x^2+4x+4-2\left(x^2+5x+6\right)+x^2+10x+25=7\)
\(\Leftrightarrow4x+10=0\Leftrightarrow x=-\frac{5}{2}\)
4 tháng 8 2021
Bạn áp dụng hằng đẳng thức số 1, nhân phá ngoặc là Ok nhé
\(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)
\(\Leftrightarrow x^2+4x+4-2\left(x^2+3x+2x+6\right)+x^2+10x+25-7=0\)
\(\Leftrightarrow2x^2+14x+22-2x^2-6x-4x-12=0\)
\(\Leftrightarrow4x+10=0\)
\(\Leftrightarrow4x=-10\)
\(\Leftrightarrow x=\frac{-5}{2}\)
29 tháng 7 2021
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
29 tháng 7 2021
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
`a,`
`x*3+7=16`
`=>3x = 16 - 7`
`=> 3x = 9`
`=> x = 9 \div 3`
`=> x = 3`
Vậy, `x = 3`
`b,`
` x- 152 \div 2 = 46?`
`=> x - 76 = 46`
`=> x = 46 + 76`
`=> x = 122`
Vậy, `x = 122.`
`c,`
` 74-2*(x+3)=34`
`=> 2(x + 3) = 74 - 34`
`=> 2(x+2) = 40`
`=> x + 2 = 40 \div 2`
`=> x + 2 = 20`
`=> x = 20 - 2`
`=> x = 18`
Vậy, `x = 18.`