1.TÌM X
A, 2X-3=5X.7
B, -4X-3X=X-5
Tìm x biết :
a, 4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6(x + 14)
b, 5.(3x + 5) - 4.(2x - 3) = 5x + 3.(2x + 12) + 1
c, 2.(5x - 8) - 3.(4x - 5) = 4.(3x - 4) + 11
d, (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
e, (8x - 3)(3x + 2) - (4x + 7)(x + 4)= (2x + 1)(5x - 1) - 33
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
b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1
=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
=> (15x - 8x) + (25 + 12) = 11x + 37
=> 7x + 37 = 11x + 37
=> 11x - 7x = 0
=> x = 0
Tìm x biết:
a) 4(18-5x) - 12(3x-7) = 15(2x-16) - 6(x+14)
b) 5(3x+5) - 4(2x-3) = 5x+3(2x+12)+1
c)2(5x-8) - 3(4x-5) = 4(3x-4) + 11
d) 5x-3 {4x-2 [4x-3(5x-2)]}=182
3,26 + 4/5 =?
làm nhanh lên giúp mình nhé
3,26+4/5=3,26+0,8=3,34
K cho mình cái
\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x+16+11\)
\(\Leftrightarrow-2x-1=12x+27\Leftrightarrow-14x-28=0\Leftrightarrow x=-2\)
Tìm x:
a) (3x+2).(x-1)-(x+2).(3x+1)=7
b) (6x+5).(2x+3)-(4x+3).(3x-2)=8
c) 2x.(x+3)-(x+1).(2x+1)-5=9
d) (5x+3).(4x-7)-(10x+9).(2x-3)=10
a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7
<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7
<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7
<=> -8x - 4 = 7
<=> -8x = 11
<=> x = -11/8
b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8
<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8
<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8
<=> 27x + 21 = 8
<=> 27x = -13
<=> x = -13/27
c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9
<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9
<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9
<=> 3x - 6 = 9
<=> 3x = 15
<=> x = 5
d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10
<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10
<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10
<=> -11x + 6 = 10
<=> -11x = 4
<=> x = -4/11
a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)
b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)
c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)
d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
Tìm x:
a) 4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
b) 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1
c) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) +11
d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Leftrightarrow156-56x=24x-324\)
\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)
\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)
\(\Leftrightarrow7x+37=11x-35\)
\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)
\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)
\(\Leftrightarrow5x+33x-18-182=0\)
\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)
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1. TÌm x:
a)4x^2-2x+3-4x.(x-5)=7x-3
b)-3x.(x-5)+5.(x-1)+3x^2=4x
c)7x.(x-2)-5.(x-1)=21x^2-14x^2+3
d)3.(5x-1)-x.(x-2)+x^2-13x=7
e) 1/5x.(10x-15)-2x.(x-5)=12
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~
1. TÌm x:
a)4x^2-2x+3-4x.(x-5)=7x-3
b)-3x.(x-5)+5.(x-1)+3x^2=4x
c)7x.(x-2)-5.(x-1)=21x^2-14x^2+3
d)3.(5x-1)-x.(x-2)+x^2-13x=7
e) 1/5x.(10x-15)-2x.(x-5)=12
a) 4x2 - 2x + 3 - 4x(x - 5) = 7x - 3
=> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
=> 18x + 3 = 7x - 3
=> 18x - 7x = -3 - 3
=> 11x = -6
=> x = -6/11
b) -3x(x - 5) + 5(x - 1) + 3x2 = 4x
=> -3x2 + 15x + 5x - 5 + 3x2 = 4x
=> 20x - 5 = 4x
=> 20x - 4x = 5
=> 16x = 5
=> x = 5/16
\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)
\(\Leftrightarrow7x^2-7x^2-19x=3-5\)
\(\Leftrightarrow-19x=-2\)
\(\Leftrightarrow x=\frac{2}{19}\)
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
<=> 18x + 3 = 7x - 3
<=> 18x = 7x - 3 - 3
<=> 18x = 7x - 6
<=> 18x - 7x = -6
<=> 11x = -6
<=> x = -6/11
=> x = -6/11
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
<=> 20x - 5 = 4x
<=> 20x = 4x + 5
<=> 20x - 4x = 5
<=> 16x = 5
<=> x = 5/16
=> x = 5/16
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
<=> 7x.(x - 2) - 5.(x - 1) = 7x2 + 3
<=> 7x2 - 19x + 5 = 7x2 + 3
<=> 7x2 - 19x = 7x2 + 3 - 5
<=> 7x2 - 19x = 7x2 - 2
<=> 7x2 - 19x - 7x2 = -2
<=> -19x = -2
<=> x = 2/19
=> x = 2/19
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
<=> 4x - 3 = 7
<=> 4x = 7 + 3
<=> 4x = 10
<=> x = 10/4
=> x = 5/2
e) 1/5x.(10x - 15) - 2x.(x - 5) = 12
<=> x(2x - 3) - 2x(x - 5) = 12
<=> 7x = 12
<=> x = 12/7
=> x = 12/7
Tìm x:
A/ 5x²(3x-1)-10x³(1-3x) = 0
B/ 9x(x-3)+18x²(3-x) - 81x²(x-3) = 0
C/ (4x-1)(3x-5)-x(4x-1) = 3(1-4x)
D/ 2x²(3x+1)+7(3x+1) = 2x(3x+1)
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x:
a) 4x(3x-7)-6(2x^2-5x+1)=12
b) (5x+3)(4x-1)+(10x-7)(-2x+3)=27
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
a) 4x(3x-7)-6(2x2-5x+1)=12
=>4x.3x-4x.7-6.2x2-6.(-5x)-6.1=12
=>12x2-28x-12x2+30x-6=12
=>2x-6 =12
=>2x =12+6
=>2x =18
=>x =18:2
=>x =6
b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27
=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27
=>20x2-5x+12x-3-20x2+30x+14x-21=27
=>39x-36 =27
=>39x =27+36
=>39x =63
=>x =63:39
=>x =21/13
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17
=>24x2+16x-15x-10-24x2+12x+14x-7=17
=>27x-17 =17
=>27x =17+17
=>27x =34
=>x =34:27
=>x =34/27
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
=>30x2-5x+63x-9 - 30x2-2x-45x-3=-190
=>11x-12 =-190
=>11x =-190+12
=>11x =-178
=>x = -178:11
=>x =-178/11
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^