Giúp mình vs:
Bài 1: hãy tìm số bị thiếu:
Câu 1:
2x-1/3 = 3x+1/4
câu 2:
3x/5x+4 = 12/28
câu 3:
x/2 = 8/x
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\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84
<=> -80x = -480
<=> x = 6
b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1
<=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0
<=> -4x = 0
<=> x = 0
c) 2(5x-8)-3(4x-5)=4(3x-4)+11
= 10x - 16 - 12x + 15 = 12x - 16 + 11
= -14x = -4
= x =\(\frac{2}{7}\)
d) 5x-3{4x-2[4x-3(5x-2)]}=182
= 5x - 3 . [4x - 2(4x - 15x + 6)]
= 5x - 3 . (4x - 8x + 30x - 12)
= 5x - 12x + 24x - 90x + 36
= -73x + 36 = 182
=> -73x = 182 - 36 = 146
=> x = 146 : (-73) = -2
~Hok tốt~
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
Câu 1:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\frac{8x-4}{12}=\frac{9x+3}{12}\)
\(\Rightarrow8x-4=9x+3\)
\(8x-9x=3+4\)
\(-x=7\Rightarrow x=-7\)
Câu 2:
\(\frac{3x}{5x+4}=\frac{12}{28}\)
\(\Rightarrow\left(3x\right).28=12.\left(5x+4\right)\)
\(84x=60x+48\)
\(84x-60x=48\)
\(24x=48\)
\(\Leftrightarrow x=2\)
Câu 3 :
\(\frac{x}{2}=\frac{8}{x}\)
\(\Rightarrow x.x=8.2\)
\(x^2=16\)
\(x^2=\left(\pm4\right)^2\)
\(\Rightarrow x\left\{-4;4\right\}\)
học tốt !!!
Câu 1: \(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Leftrightarrow2\left(2x-1\right)=3\left(3x+1\right)\)
\(\Leftrightarrow4x-2=9x+3\)
<=> -9x+4x=3+2
<=> -5x=5
<=> x=-1
Câu 3: \(\frac{x}{2}=\frac{8}{x}\left(x\ne0\right)\)
<=> x2=16
<=> \(\orbr{\begin{cases}x=4\\x=-4\end{cases}\left(tmđk\right)}\)
Câu 2: Bạn cũng nhân chéo lên nhưng nhớ đi điều kiện: \(x\ne\frac{-4}{5}\)