Câu 1:tìm x , biết
a,150:[17-(2x-3)]=350
b, 4.[105-(x-9)]-486=0
c,1000-2{600- 4.[198-7.(21-7)]}.x=200
Câu 2:Tìm x, biết
a,350:[19-(7x-3)]=350
b, 7.[109-(x-9)]-420=0
c,254-2{200-4.[90-7.(19-9)]}.x=54
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\(\text{a)}350:\left[19-\left(7x-3\right)\right]=350\)
\(\Rightarrow19-\left(7x-3\right)=1\)
\(\Rightarrow19-7x+3=1\)
\(\Rightarrow-7x=-21\)
\(\Rightarrow x=3\)
\(\text{b)}7.\left[109-\left(x-9\right)\right]-420=0\)
\(\Rightarrow7\left(109-x+9\right)=420\)
\(\Rightarrow118-x=60\)
\(\Rightarrow x=58\)
\(\text{c)}252-2\left\{200-4\left[90-7.\left(19-9\right)\right]\right\}.x=4\)
\(\Rightarrow2\left\{200-4\left[90-7.10\right]\right\}.x=250\)
\(\Rightarrow\left[200-4\left(90-70\right)\right].x=125\)
\(\Rightarrow\left[200-4.20\right].x=125\)
\(\Rightarrow120x=125\)
\(\Rightarrow x=\frac{125}{120}=\frac{25}{24}\)
\(\text{Xin điểm ạ}\)
350:[19-(7x-3)]=350
[19-(7x-3)]=350:350
[19-(7x-3)]=1
(7x-3)=19-1
(7x-3)=18
7x=18+3
7x=21
x=21:7
x=3
vậy x=3
7.[109-(x-19)]-420=0
7.[109-(x-19)]=0+420
7.[109-(x-19)]=420
[109-(x-19)]=420:7
[109-(x-19)]=60
(x-19)=109-60
(x-19)=49
x=49+19
x=58
vậy x=58
254-2{200-4.[90-7.(19-9)]}.x=4
254-2{200-4.[90-7.10]}.x=4
254-2{200-4.[90-70]}.x=4
254-2{200-4.20}.x=4
254-2{200-80}.x=4
254-2.120.x=4
254-240.x=4
240.x=254-4
240.x=250
x=250:240
x=25/24
vậy x=25/24
Bài 2:
a: =>x-1=1 hoặc x-1=-1
=>x=2 hoặc x=0
b: =>x+1=-1
hay x=-2
c: =>(135-7x):9=8
=>135-7x=72
=>7x=63
hay x=9
d: =>(x+7)(x-3)<0
=>-7<x<3
e: \(\Leftrightarrow3^{x-3}=18+9=27\)
=>x-3=3
hay x=6
f: =>4-2x=0
hay x=2
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\Rightarrow x=-2\\ b,\Rightarrow\left(x-2021\right)\left(2022x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{2022}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-9\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-7\right)=0\\ \Rightarrow\left(x-3\right)\left(-4-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
1:
a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)
b:
Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\)
\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)
có rút gọn phân số 35/6 về phân số đc ko ? Nếu có thì mn rút gọn cho mik nhé !
Ai xong tr tui k cho
a) 150 : [17 - (2x - 3)] = 350
=> 17 - (2x - 3) = \(\frac{150}{350}=\frac{3}{7}\)
=> 2x - 3 = \(17-\frac{3}{7}=\frac{116}{7}\)
=> 2x = \(\frac{116}{7}+3\)
=> 2x = \(\frac{137}{7}\)
=> x = \(\frac{137}{7}:2=\frac{137}{14}\)
b) 4 . [105 - (x - 9)] - 486 = 0
=> 4. [105 - (x - 9)] = 486
=> 105 - (x - 9) = 243/2
=> x - 9 = \(105-\frac{243}{2}=-\frac{33}{2}\)
=> x = \(-\frac{33}{2}+9=-\frac{15}{2}\)
c) 1000 - 2{600 - 4.[198 - 7.(21 - 7)]}.x = 200
=> 1000 - 2{600 - 4.[198 - 7.14]}.x = 200
=> 1000 - 2{600 - 4.[198 - 98].x = 200
=> 1000 - 2{600 - 4.100}.x = 200
=> 1000 - 2{600 -400}.x = 200
=> 1000 - 2.200.x = 200
=> 2.200.x = 800
=> 400.x = 800
=> x = 2
Câu 2 :
a) 350 : [19 - (7x - 3)] = 350
=> 19 - (7x - 3) = 1
=> 7x - 3 = 18
=> 7x = 21
=> x = 3
b) 7.[109 - (x - 9)] - 420 = 0
=> 7.[109 - (x - 9)] = 420
=> 109 - (x - 9) = 60
=> x - 9 = 49
=> x = 49 + 9 = 58
c) 254 - 2{200 - 4.[90 - 7.(19 - 9)]}.x = 54
=> 254 - 2{200 - 4.[90 - 7.10]}.x = 54
=> 254 - 2{200 - 4[90 - 70]} . x= 54
=> 254 - 2{200 - 4.20} . x = 54
=> 254 - 2{200 - 80}.x = 54
=> 254 - 2.120 . x= 54
=> 2.120.x = 200
=> 240.x = 200
=> \(x=\frac{5}{6}\)