a) 13-(40-x)=35
b) 14-3*(5-x)=8
c) (3x-2)*(5+x)=0
d) (3x-6)*3=3^4
K
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Những câu hỏi liên quan
6 tháng 1 2023
b)x+3=4:2
=> x=-1
d)5x-15=3x-5
<=> 5x-3x=15-5
<=> 2x=10
<=> x=5
f) 35-7x=11-5x
<=> 35-11=-5x+7x
<=> 24=2x
<=> x=12
6 tháng 1 2023
h) 6x-2-3x=10
<=> 3x=10+2
<=> x=4
j)3-2x=3x+3-x-2
<=> 3-2x=2x+1
<=>-4x=-2
<=> x=1/2
9 tháng 9 2021
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
30 tháng 10 2021
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
LM
24 tháng 3 2023
\(a,x+\dfrac{5}{7}=\dfrac{10}{-35}\\ x=\dfrac{-10}{35}-\dfrac{5}{7}\\ x=-1\\ b,\dfrac{2}{x}=\dfrac{x}{8}\\ x.x=2.8\\ x^2=16\\ x=\pm4\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 3 2023
a, \(x\) + \(\dfrac{5}{7}\) = \(\dfrac{10}{-35}\)
\(x\) = \(-\dfrac{10}{35}\) - \(\dfrac{5}{7}\)
\(x\) = -1
b, \(\dfrac{2}{x}\) = \(\dfrac{x}{8}\)
\(x^2\) = 16
\(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -4; 4}
c, \(\dfrac{x-3}{2}\) = \(\dfrac{32}{x-3}\)
(\(x-3\))(\(x-3\)) = 32\(\times\) 2
(\(x-3\))(\(x\) - 3) = 64
(\(x-3\))2 = 82
\(\left[{}\begin{matrix}x-3=-8\\x-3=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-8+3\\x=8+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=11\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -5; 11}
27 tháng 1 2021
a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)
\(\Leftrightarrow3x+2x-10=6-5x+1\)
\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm
b, \(x^3-3x^2-x+3=0\)
\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)
Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }
27 tháng 1 2021
c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3+x^2-3x-2=0\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn
Vậy ...
LL
30 tháng 4 2021
a. 2x\(^2\)-8=0
2x\(^2\)=8
x\(^2\)=4
x=2
b.3x\(^3\)-5x=0
x(3x\(^2\)-5)=0
\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)
LL
1 tháng 5 2021
c.x\(^4\)+3x\(^2\)-4=0\(^{\left(\cdot\right)}\)
đặt t=x\(^2\) (t>0)
ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)
thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm
t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4
khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1
khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2
vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2
d)3x\(^2\)+6x-9=0
thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm
x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)
e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\) (ĐK: x#5; x#2 )
⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)
⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0
⇔-7x\(^2\) - 6x + 46=0
Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0
\(\sqrt{\Delta'}=\sqrt{62}\)
vậy pt có 2 nghiệm phân biệt
x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)
x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)
vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......
câu g làm tương tự câu c
15 tháng 4 2022
a: x<5 thì 5-x>0
A=5x+5-x+5=4x+10
b: Khi x>=0 thì \(B=5x+10+3x=8x+10\)
Khi x<0 thì B=5x+10-3x=2x+10
d: Khi x>=3 thì \(D=x-3-3x+15=-2x+12\)
Khi x<3 thì D=3-x-3x+15=-4x+18
15 tháng 7 2022
Bài 1:
8: \(=\dfrac{x+3}{x\left(x-3\right)}\)
9: \(=\dfrac{x-2}{x-5}\cdot\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x-2\right)^2}=\dfrac{x+5}{x-2}\)
10: \(=1:\dfrac{a-1}{a}=\dfrac{a}{a-1}\)
12: \(=\dfrac{6\left(x+1\right)}{3x\left(x+1\right)}=\dfrac{2}{x}\)
13: \(\dfrac{3}{x+3}-\dfrac{x-6}{x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2x+6}{x\left(x+3\right)}=\dfrac{2}{x}\)
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
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\(a)13-\left(40-X\right)=35\)
\(\Leftrightarrow40-X=13-35\)
\(\Leftrightarrow40-X=-22\)
\(\Leftrightarrow X=40-\left(-22\right)\)
\(\Leftrightarrow X=62\)
Vậy \(X=62\)
\(b)14-3x\left(5-X\right)=8\)
\(\Leftrightarrow3x\left(5-X\right)=14-8\)
\(\Leftrightarrow3x\left(5-X\right)=6\)
\(\Leftrightarrow5-X=6:3\)
\(\Leftrightarrow5-X=3\)
\(\Leftrightarrow X=5-3\)
\(\Leftrightarrow X=2\)
Vậy \(X=2\)
\(c)\left(3X-2\right)x\left(5+X\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3X-2=0\\5+X=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3X=0+2\\X=0-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3X=2\\X=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X=2:3\\X=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X=\frac{2}{3}\\X=-5\end{cases}}\)
Vậy \(X\in\left\{-5;\frac{2}{3}\right\}\)
\(d)\left(3X-6\right)x3=3^4\)
\(\Leftrightarrow\left(3X-6\right)x3=81\)
\(\Leftrightarrow3X-6=81:3\)
\(\Leftrightarrow3X-6=27\)
\(\Leftrightarrow3X=27+6\)
\(\Leftrightarrow3X=33\)
\(\Leftrightarrow X=33:3\)
\(\Leftrightarrow X=11\)
Vậy\(X=11\)