(3x-2)3 = 2 . 25
5x+1-5x= 500
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a: Ta có: \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)
\(=142-\left[50-80+40\right]\)
=142-10
=132
a. 142 - [50-(23.10-23.5)]
= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]
= 142 - [ 50 - ( 80 - 40 )]
= 142 - [ 50 - 40 ]
= 142 - 10 = 132
b. 375 : { 32[ 4+(5.32 - 42 )]} - 14
= 375 : { 32[4+(5.9 - 42 )]} - 14
= 375 : { 32[4 + ( 45 - 42 )]} - 14
= 375 : {32[4+3]} - 14
= 375 : 224 - 14
c.{210 : [ 16+3.(6+3.2^2)]} - 3
= {210 : [ 16 + 3.(6+3.4)]} - 3
= { 210 :[16+3.(6+12)]}-3
= {210 : [ 16+3.18)]} - 3
= { 210 : [ 16 + 54]} - 3
= { 210 : 70 } - 3
= 30 - 3 = 27
d.500-{5.[409-(2^3 . 3-21^2)] - 1724}
= 500-{5.[409-(8 . 63 - 21^2] - 1724}
= 500 - { 5.[409- (504 - 441) - 1724}
= 500 -{ 5.[ 409 - 63 ] - 1724}
= 500 - { 5.346 - 1724}
= 500 - { 1720 - 1724 }
= 500 - 6
= 494
1) 5x = 120 : 2
5x = 60
x = 60 : 5
x = 12.
2) 3x + 27 = 45
3x = 45 - 27
3x = 18
x = 18 : 3
x = 6.
3) 73 + x - 122 = 234 : 2
73 + x - 122 = 117
73 + x = 117 + 122
73 + x = 239
x = 239 - 73
x = 166.
4) 32 - 2 ( x - 1 ) = 12
2 ( x - 1 ) = 32 - 12
2 ( x - 1 ) = 20
x - 1 = 20 : 2
x - 1 = 10
x = 10 + 1
x = 11.
5) 3 ( x + 1 ) - 15 = 36
3 ( x + 1 ) = 36 + 15
3 ( x + 1 ) = 51
x + 1 = 51 : 3
x + 1 = 17
x = 17 - 1
x = 16.
6) 32 - ( 4x - 16 ) = 21
4x - 16 = 32 - 21
4x - 16 = 11
4x = 11 + 16
4x = 27
x = 27 : 4
x = 6,75.
7) 13x - 3x = 500
( 13 - 3 )x = 500
10x = 500
x = 500 : 10
x = 50.
8) 9x + x - 35 = 68
9x + 1x - 35 = 68
9x + 1x = 68 + 35
9x + 1x = 103
( 9 + 1 )x = 103
10x = 103
x = 103 : 10
x = 10,3.
1) 5x = 120 : 2
5x = 60
x = 60 : 5
x = 12.
2) 3x + 27 = 45
3x = 45 - 27
3x = 18
x = 18 : 3
x = 6.
3) 73 + x - 122 = 234 : 2
73 + x - 122 = 117
73 + x = 117 + 122
73 + x = 239
x = 239 - 73
x = 166.
4) 32 - 2 ( x - 1 ) = 12
2 ( x - 1 ) = 32 - 12
2 ( x - 1 ) = 20
x - 1 = 20 : 2
x - 1 = 10
x = 10 + 1
x = 11.
5) 3 ( x + 1 ) - 15 = 36
3 ( x + 1 ) = 36 + 15
3 ( x + 1 ) = 51
x + 1 = 51 : 3
x + 1 = 17
x = 17 - 1
x = 16.
6) 32 - ( 4x - 16 ) = 21
4x - 16 = 32 - 21
4x - 16 = 11
4x = 11 + 16
4x = 27
x = 27 : 4
x = 6,75.
7) 13x - 3x = 500
( 13 - 3 )x = 500
10x = 500
x = 500 : 10
x = 50.
8) 9x + x - 35 = 68
9x + 1x - 35 = 68
9x + 1x = 68 + 35
9x + 1x = 103
( 9 + 1 )x = 103
10x = 103
x = 103 : 10
x = 10,3.
a) 5x = 120 : 2
5x = 60
x = 60 : 5
x = 12
b) 3x + 27 = 45
3x = 45 - 27
3x = 18
x = 18 : 3
x = 6
c) 73 + x - 122 = 234 : 2
73 + x - 122 = 117
73 + x = 117 + 122
73 + x = 239
x = 239 - 73
x = 166
d) 32 - 2(x - 1) = 12
2(x - 1) = 32 - 12
2(x - 1) = 20
x - 1 = 20 : 2
x - 1 = 10
x = 10 + 1
x = 11
e) 3(x + 1) - 15 = 36
3(x + 1) = 36 + 15
3(x + 1) = 51
x + 1 = 51 : 3
x + 1 = 17
x = 17 - 1
x = 16
f) 21 - (4x - 16) = 21
4x - 16 = 21 - 21
4x - 16 = 0
4x = 0 + 16
4x = 16
x = 16 : 4
x = 4
g) 13x - 3x = 500
10x = 500
x = 500 : 10
x = 50
h) 9x + x - 35 = 65
10x - 35 = 65
10x = 65 + 35
10x = 100
x = 100 : 10
x = 10
a: =>|5x-2|=|2x-3|
=>5x-2=2x-3 hoặc 5x-2=-2x+3
=>3x=-1 hoặc 7x=5
=>x=5/7 hoặc x=-1/3
b: =>|5x-2|-|2x+2|=3x+5
TH1 x<-1
PT sẽ là 2-5x+2x+2=3x+5
=>-3x+4=3x+5
=>-6x=1
=>x=-1/6(loại)
TH2: -1<=x<2/5
Pt sẽ là 2-5x-2x-2=3x+5
=>-7x=3x+5
=>-4x=5
=>x=-5/4(loại)
Th3: x>=2/5
PT sẽ là 5x-2-2x-2=3x+5
=>3x-4=3x+5
=>0x=9(loại)
Tìm min:
$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$
$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$
$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$
Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$
Tìm min
$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$
$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)
Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$
$\Leftrightarrow x=\frac{-1}{4}$
a, 3 x - 2 3 = 2 . 32
3 x - 2 3 = 64
3 x - 2 3 = 4 3
3 x - 2 = 4 ⇔ x = 2
b, 5 x + 1 - 5 x = 500
5 x . 5 - 5 x = 500
5 x . ( 5 - 1 ) = 500
5 x . 4 = 500
5 x = 125 = 5 3
x = 3
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(TH_1:3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(TH_2:-2x-7=0\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(TH_1:x=0\)
\(TH_2:x-1=0\)
\(\Leftrightarrow x=1\)
\(TH_3:2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)
\(TH_1:3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
\(TH_2:2x-4=0\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Rightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x-9x=-6-16+12\)
\(\Leftrightarrow11x=-10\)
\(\Leftrightarrow x=-\dfrac{10}{11}\)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow3x+1=5x+8\)
\(\Leftrightarrow3x-5x=8-1\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(X=\dfrac{-7}{2}\)
b) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow9x^2-16-3x^2-4x=0\)
\(\Leftrightarrow6x^2-4x-16=0\)
\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)
\(\Leftrightarrow3x^2-6x+4x-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Leftrightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x+16-12-9x+6=0\)
\(\Leftrightarrow11x+10=0\)
\(\Leftrightarrow x=\dfrac{-10}{11}\)
Vậy \(x=\dfrac{-10}{11}\)
\(5^{x+1}-5^x=500\)
\(5^x.5-5^x=500\)
\(5^x.\left(5-1\right)=500\)
\(5^x.4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow x=3\)
( 3 x - 2 ) 3 = 2 . 2 5
( 3 x - 2 ) 3 = 64
( 3 x - 2 )3 = 4 3
=> 3 x - 2 = 4
3 x = 6
x = 2
5 x + 1 - 5 x = 500
5 x . ( 5 - 1 ) = 500
5 x . 4 = 500
5 x = 125
5 x = 5 3
=> x = 3