Tìm x:
x:1/2+x:1/4+x:1/6+x:1/8+x:1/10=1
Tìm x:
x - 1/8 = 1/4 x 2/3
`x-1/8=1/4xx2/3`
`x-1/8=1/6`
`x=1/6+1/8`
`x=7/24`
x - 1/8 = 1/4 x 2/3
x - 1/8= 1/6
x = 1/6 + 1/8
x = 14/48
1. Tính rồi rút gọn:
a) (x - 7)(x + 7) - x^2
2. Tìm x:
x(x - 4) - x^2 + 8 = 0
Bài 2:
Ta có: \(x\left(x-4\right)-x^2+8=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2
1)=x2-49-x2
=-49
2)=>x2-4x-x2+8=0
=>-4x+8=0
=>-4x=-8
=>x=2
Tìm X:x/2+x-1/3+x-2/4=1/4
Tìm x:x+1/2=8/x+1
\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)
=>(x+1)2=16
=>x+1=\(\sqrt{16}\)
=>x+1=4
=>x=3
tìm x:
X:1/3=1/2+1/3 (X:2/3):2/5=7/10
x : 1/3 = 1/2 +1/3
=>x:1/3=5/6
=>x=5/6 x 1/3
=>x=5/18
(x:2/3):2/5=7/10
=>x:2/3=7/10 x 2/5
=>x:2/3=7/25
=>x=7/25x2/3
=>x=14/75
CHÚC BẠN HỌC TỐT
tìm x:
X:1/3=1/2+1/3 (X:2/3):2/5=7/10
x:1/3=1/2+1/3
x:1/3=5/6
x =5/6.1/3
x =5/18
(x:2/3):2/5=7/10
x:2/3 =7/10.2/5
x:2/3 =7/25
x:2/3 =7/25.2/3
x:2/3 =14/75
Tìm x:
x(x-1)(x+3)-x^2(x+3)=-4
\(\Leftrightarrow x^3+2x^2-3x-x^3-3x^2=-4\)
\(\Leftrightarrow x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>x=-4 hoặc x=1
bài 1: Tìm X:x:8= 35 (dư 4)
; x:8= 42 (dư 6)
; x:8= 35 (số dư nhỏ nhất )
; x:8= 35 (số dư)
x:8=35( dư 4)
x= (35x8)+4
x=.....
Các bài khác tương tự.
Số dư trong bài x:8= 35( số dư nhỏ nhất bằng 7)
dễ mà mk làm 1 câu mẫu cho bn nhé còn các câu sau tương tự
x : 8 = 35 ( du 4 )
x = 35 * 8 + 4
x = 284
k mk nha
các bạn giải hộ rõ ra cho mik với ...c.ơn nhiều
Bài 1 :Phân tích đa thức sau thành nhân tử
(12x2+6x)(y+z)+(12x2+6x)(y-z)
Bài 2:tìm x:
x(x-6)+10(x-6)=0
1.
\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)
2.
\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt
Bài 1:
Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=6x\left(2x+1\right)\cdot2y\)
\(=12xy\left(2x+1\right)\)
Bài 2:
Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
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) ⇒x=56−12=56−36=26=13⇒x=56−12=56−36=26=13
b) ⇒x=34−14=24=12⇒x=34−14=24=12
c) ⇒x=310+15=310+210=510=12⇒x=310+15=310+210=510=12
d) ⇒x=56−13=56−26=36=12⇒x=56−13=56−26=36=12
e) ⇒x=12−310=510−310=210=15⇒x=12−310=510−310=210=15
g) ⇒x=38−14=38−28=18⇒x=38−14=38−28=18
Đọc tiếp
a) ⇒x=56−12=56−36=26=13⇒x=56−12=56−36=26=13
b) ⇒x=34−14=24=12⇒x=34−14=24=12
c) ⇒x=310+15=310+210=510=12⇒x=310+15=310+210=510=12
d) ⇒x=56−13=56−26=36=12⇒x=56−13=56−26=36=12
e) ⇒x=12−310=510−310=210=15⇒x=12−310=510−310=210=15
g) ⇒x=38−14=38−28=18⇒x=38−14=38−28=18