Tìm x
(x+3)^3 : 3-1= -10
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{1}{2}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
b) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c) \(x-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
d) \(\dfrac{5}{6}-x=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
e) \(\dfrac{3}{10}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{10}=\dfrac{5}{10}-\dfrac{3}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
g) \(x+\dfrac{1}{4}=\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{3}{8}-\dfrac{2}{8}=\dfrac{1}{8}\)
a) 12+x=5612+x=56
⇒x=56−12=56−36=26=13⇒x=56−12=56−36=26=13
b) x+14=34x+14=34
⇒x=34−14=24=12⇒x=34−14=24=12
c) x−15=310x−15=310
⇒x=310+15=310+210=510=12⇒x=310+15=310+210=510=12
d) 56−x=1356−x=13
⇒x=56−13=56−26=36=12⇒x=56−13=56−26=36=12
e) 310+x=12310+x=12
⇒x=12−310=510−310=210=15⇒x=12−310=510−310=210=15
g) x+14=38x+14=38
⇒x=38−14=38−28=18⇒x=38−14=38−28=18
Đọc tiếp
a: =>10+3x-3=6x+10
=>3x-3=6x
=>-3x=3
=>x=-1
b: =>x+1=0 hoặc x-2=0
=>x=-1 hoặc x=2
a) \(10+3\left(x-1\right)=10+6x\)
\(\Rightarrow10+3x-3=10+6x\)
\(\Rightarrow3x-6x=10-10+3\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-\dfrac{3}{3}\)
\(\Rightarrow x=-1\)
b) \(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
1.Thử với p=2 ko được.
Thử với p=3 được.
Thử với >3 là 3k+1 và 3k+2.
=>p=3.
2.Phân tích 10 ra thừa số nguyên tố rồi kẻ bảng.
3.Tách riêng x và số ra.
Câu 1 : \(\frac{10}{3}\cdot x-\frac{1}{2}=\frac{1}{10}\)
\(\Rightarrow\frac{10}{3}\cdot x=\frac{3}{5}\Rightarrow x=\frac{9}{50}\)
Câu 2 : \(\frac{1}{3}\cdot x+\frac{1}{3}\cdot x=\frac{1}{8}\Rightarrow x\cdot\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{1}{8}\)
\(x\cdot\frac{2}{3}=\frac{1}{8}\Rightarrow x=\frac{3}{16}\)
Câu 3 : \(\left[3x-1\right]\left[\frac{1}{2}x-\frac{2}{5}\right]=0\)
\(\Rightarrow3x-1=0\)hoặc \(\frac{1}{2}x-\frac{2}{5}=0\)
( vô lí ) x = 4/5
\(\left(x+3\right)^3:3-1=\left(-10\right)\)
\(\left(x+3\right)^3:3=\left(-10\right)+1\)
\(\left(x+3\right)^3:3=\left(-9\right)\)
\(\left(x+3\right)^3=\left(-9\right):3\)
\(\left(x+3\right)^3=\left(-3\right)\)
\(\left(x+3\right)^3=\left(-1\right)^3\)
\(x+3=\left(-1\right)\)
\(x=\left(-1\right)-3\)
\(x=\left(-4\right)\)
@Nghệ Mạt
#cua