35+3.(2x+4)=-1
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23 tháng 2 2023
a: =>31-x=60
=>x=-29
b: =>(x-140):35=280-270=10
=>x-140=350
=>x=490
c: =>(1900-2x):35=48
=>1900-2x=1680
=>2x=220
=>x=110
d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)
=>2x-1=11
=>x=6
e: =>(x+2)^5=4^5
=>x+2=4
=>x=2
f: =>3x-4=0 hoặc x-1=0
=>x=4/3 hoặc x=1
g: =>(2x-1)^2=49
=>2x-1=7 hoặc 2x-1=-7
=>x=-3 hoặc x=4
h: =>x(x+1)/2=78
=>x(x+1)=156
=>x=12
CT
1
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
4 tháng 4 2020
1 ) ( 2x - 15 ) + 17 = 6
=> 2x - 15 = 6 - 17
=> 2x - 15 = - 11
=> 2x = - 11 + 15
=> 2x = 4
=> x = 4: 2
=> x = 2
Vậy x = 2
2) 15 - ( 4 - 3x ) = -4
=> 4 - 3x = 15 - ( -4 )
=> 4 - 3x = 19
=> 3x = 4 - 19
=> 3x = -15
=> x = -15 : 3
=> x = -5
Vậy x = -5
3) -21 + 3 . ( -x + 7 ) = - 18
=> 3 . ( -x + 7 ) = ( -18 ) - ( -21 )
=> 3 . ( -x + 7 ) = 3
=> -x +7 = 3 : 3
=> -x + 7 = 1
=> -x = 1 - 7
=> -x = - 6
=> x = - (- 6 )
Vậy x = - ( -6 )
4 ) 78 : ( 3x - 2 ) = -3
=> 3x - 2 = 78 : ( -3)
=> 3x - 2 = -26
=> 3x = -26 + 2
=> 3x = -24
=> x = -24 : 3
=> x = -8
Vậy x = -8
5) -35 : 5 . ( -3 - 2x ) = 35
=> -7 . ( -3 ) + 2x = 35
=> -3 + 2x = 35 : ( -7)
=> -3 + 2x = -5
=> 2x = (-5) - (-3)
=> 2x = -2
=> x = -2 : 2
=> x = -
Vậy x = -1
~ Chúc bạn học tốt ~
4 tháng 4 2020
1/ (2x – 15) + 17 = 6
<=> 2x=4
<=> x=2
2/ 15 – (4 – 3x) = - 4
<=> 3x=-15
<=> x=-5
3/ - 21 + 3(-x + 7) = -18
<=> -3x=-18
<=> x=6
4/ 78 : (3x – 2) = -3
<=> 3x-2=-26
<=> 3x=-24
<=> x=-8
5/ -35 : 5.(-3 – 2x) = 35
<=> -3-2x=-7
<=> 2x=4
<=> x=2
MV
8 tháng 8 2017
\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
6 tháng 8 2017
280 - ( x - 140 ) : 35 = 270
<=> ( x - 140 ) : 35 = 280 -270
<=> ( x -140 ) : 35 = 10
<=> x -140 = 10 . 35
<=> x -140 = 350
<=> x = 350 + 140
<=> x = 490
TP
6 tháng 8 2017
1) 280 - ( x - 140 ) : 35 = 270
=> ( x - 140 ) : 35 = 280 - 270 = 10
x - 140 = 350
=> x = 350 + 140
=> x = 390
2) ( 190 - 2x ) : 35 - 32 = 16
190 - 2x = ( 16 + 32 ) . 35 = 1680
x = ( 190 - 1680 ) : 2
x = -745
3) 720 : { 41 - ( 2x - 5 )} -2.5
Sai đề.
4) ( x : 23 + 45 ) . 37 - 22 = 24 .105
x : 23 + 45 = ( 24.105 + 22 ) : 37
x : 23 = 2542/37 - 45 = 877/37
x = 877/37.23 = 20171/37
5) ( 3x - 4 ) ( x - 1 ) = 0
=> 3x - 4 = 0 hoặc x - 1 = 0
3x - 4 = 0 hoặc x - 1 = 0
=> x = 4/3 => x = 1
Vậy x \(\in\) { 4/3;1 }
6) 22x : 4 = 83
=> 22x = 83 . 4 = 2048 = 211
=> 2x = 11
=> x = 11/2
27 tháng 6 2021
b) Ta có: \(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)
\(=x^3+2x^2-7x-14-\left(2x^2-28x-x+14\right)+x^3-2x^2-22x+35\)
\(=2x^3-29x+21-2x^2+29x-14\)
\(=2x^3-2x^2+7\)
12 tháng 7 2021
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
35+3.(2x+4)=-1
= 3.(2x+4) = -1 - 35
= 3.(2x+4) = -36
= 2x+4 = -36: 3
= 2x + 4 = -12
= 2x = -12 - 4
= 2x = -16
= x = -16 : 2
= x = -8