a, ( x-7 ) : 5 = 0
b, ( x:7-7 )( x:12-12) = 0
c, 300 - x : 5 = 273
d, 135 + x : 2 = 150
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CM
20 tháng 4 2017
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
CM
12 tháng 8 2019
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
K
4
UT
8 tháng 2 2022
a) 4x – 20 = 0
⇔ 4x = 20
⇔ x = 20 : 4
⇔ x = 5
Vậy phương trình có nghiệm duy nhất x = 5.
b) 2x + x + 12 = 0
⇔ 3x + 12 = 0
⇔ 3x = -12
⇔ x = -12 : 3
⇔ x = -4
Vậy phương trình đã cho có nghiệm duy nhất x = -4
c) x – 5 = 3 – x
⇔ x + x = 5 + 3
⇔ 2x = 8
⇔ x = 8 : 2
⇔ x = 4
Vậy phương trình có nghiệm duy nhất x = 4
d) 7 – 3x = 9 – x
⇔ 7 – 9 = 3x – x
⇔ -2 = 2x
⇔ -2 : 2 = x
⇔ -1 = x
⇔ x = -1
Vậy phương trình có nghiệm duy nhất x = -1.
TT
8 tháng 2 2022
Em lớp 6 em chỉ làm dc phần a,b,c
Kết quả như sau:
a,4x-20=0
4x=20+0
4x=20
x=20:4
x=5
NA
1
SN
20 tháng 7 2021
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
VT
1
B
23 tháng 9 2020
Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x\div7-7=0\\x.3-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\div7=7\\x.3=12\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=49\\x=4\end{cases}}\)
\(b\text{)}\Leftrightarrow...\Leftrightarrow x\div2=15\Leftrightarrow x=15.2=30\)
\(c\text{)}...\Leftrightarrow100\div x=20\Leftrightarrow x=100\div20=5\)
\(d\text{)}...\Leftrightarrow x\div5=27\Leftrightarrow x=27.5=135\)
10 tháng 8 2023
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
10 tháng 8 2023
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
10 tháng 12 2021
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
NT
2
LQ
2
11 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)
\(a,\left(x-7\right):5=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
\(b,\left(x:7-7\right)\left(x:12-12\right)=0\)
\(\Rightarrow\hept{\begin{cases}x:7=7\\x:12=12\end{cases}}\)
\(\Rightarrow x=1\)
\(c,300-x:5=273\)
\(\Rightarrow x:5=27\)
\(\Rightarrow x=27.5=135\)
\(d,135+x:2=150\)
\(\Rightarrow x:2=15\)
\(\Rightarrow x=30\)
a) \(\left(x-7\right):5=0\)
\(\Rightarrow x-7=0\times5\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=0+7\)
\(\Rightarrow x=7\)
Vậy x = 7
b) \(\left(x:7-7\right)\left(x:12-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x:7-7=0\\x:12-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x:7=7\\x:12=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=49\\x=144\end{cases}}\)
Vậy x = 49 hoặc x = 144
c) \(300-x:5=273\)
\(\Rightarrow x:5=300-273\)
\(\Rightarrow x:5=27\)
\(\Rightarrow x=27\times5\)
\(\Rightarrow x=135\)
Vậy x = 135
d) \(135+x:2=150\)
\(\Rightarrow x:2=150-135\)
\(\Rightarrow x:2=15\)
\(\Rightarrow x=15\times2\)
\(\Rightarrow x=30\)
Vậy x = 30
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