Tìm x biết
a) 3x.(x-2) = 3x\(^2\) - 6x +1
b 5.(x-3) = 2x-(3x+1)
c) 3.(2x-3) = 5+2.(x-3)
d) 6.(x-3) = 5.(x+2) - 3x
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
^^
Tìm x biết
a) 3x(2x+3)-(2x+5)(3x-2)=8
b) (3x-4)(2x+1)-(6x+5)(x-3)=3
c) 2(3x-1)(2x+5)-6(2x-1)(x+2)= - 6
d) 3xy(x+y)-(x+y)(x2+y2+2xy)+y3=27
hình như lớp 8 mà mình bấm bị lộn ai bik chỉ mình vs
a) 3x( 2x + 3) -(2x+5)(3x-2)=8
<=> 6x^2+9x-6x^2+4x-15x+10=8
<=> -2x+10=8
<=> -2x= 8-10 = -2
<=> x=1
b) (3x-4)(2x+1)-(6x+5)(x-3)=3
<=> 6x^2+3x-8x-4-6x^2+18x-5x+15=3
<=> -8x+11=3
<=> -8x= -8
<=> x=1
c, 2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
<=> 2(6x^2+15x-2x-5)-6(2x^2+4x-x-2)=6
<=> 2(6x^2+13x-5)-6(2x^2+3x-2)=6
<=> 12x^2+ 26x-10-12x^2-18x+12=6
<=> 8x+2=6
<=> 8x=4
<=> x= 1/2
d, 3xy(x+y)-(x+y)(x^2 +y^2+2xy)+y^3=27
<=> 3x2y+3xy2-(x+y)(x+y)2+y3=27
<=> 3x2y+3xy2-(x+y)3+y3=27
<=> 3x2y +3xy2 -x3-3x2y-3xy2-y3+y3=27
<=> -x3=27
<=> x= \(-\sqrt[3]{27}\)= -3
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: \(x=-2\)
Mọi người giúp tới gấp nhé:
1. Tìm x, biết:
a/ 3(2x - 3) + 2(2 - x) = -3
b/ 2x(x2 - 2) + x2(1 - 2x) - x2 = -12
2. Tìm x, biết:
a/ 3x(2x + 3) - (2x + 5)(3x - 2) = 8
b/ 4x(x - 1) - 3(x2 - 5) - x2 = (x - 3) - (x + 4)
c/ 2(3x - 1)(2x + 5) - 6(2x - 1)(x + 2) = -6
d/ 3(2x - 1)(3x - 1) - (2x - 3)(9x -1) - 3 = -3
e/ (3x - 1)(2x + 7) - (x + 1)(6x - 5) = (x + 2) - (x - 5)
f/ 3xy(x + y) - (x + y)(x2 + y2 + 2xy) + y3 = 27
3. Chứng minh rằng giá trị của các biểu thức sau không phụ thuộc vào x:
a/ A = 2x(x - 1) - x(2x + 1) - (3 - 3x)
b/ B = 2x(x - 3) - (2x - 2)(x - 2)
c/ C = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
d/ D = (2x + 11)(3x - 5) - (2x + 3)(3x + 7)
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
Bài 1:
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(6x-9+4-2x=-3\)
\(4x=-2\)
\(x=-\frac{1}{2}\)
b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)
\(2x^3-4x+x^2-2x^3-x^2=-12\)
\(-4x=-12\)
\(x=\frac{1}{3}\)
Bài 2:
a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(6x^2+9x-6x^2-15x+4x+10=8\)
\(-2x=8\)
\(x=-4\)
b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)
\(4x^2-4x-3x^2+15-x^2=-7\)
\(-4x=-22\)
\(x=\frac{11}{2}\)
c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)
\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)
\(6x-4x-10-12x+6x+12=-6\)
\(-4x=-8\)
\(x=2\)
Tìm x
a.(x+2).(x+3)-(x-2).(x+5) = 0
b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)
c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)
d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
Bài 1 : Tìm x, biết
a) 3x(x2 - 2) + x2(1-2x) - x2 = -12
b) 3x(2x +3) - (2x +5)(3x -2) = 8
c)3(3x -1)(2x+5) - 6(2x -1)(x+2) = -6
d) 3(2x -1)(3x - 1)- (2x -3)(9x -1) - 3 = -3
e) 4x(x-1)- 3(x2 -5)- x2 = (x-3)(x+4)
f) (3x-1)(2x+7)-(x+1)(6x-5)= (x+2)-(x-5)
tck đầu tiên chọn câu trả lời của mình đi
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
Bài 1 : chứng minh rằng các biểu thức sau đây không phụ thuộc vào x a,A=(3x+7)(2x+3)-(2x+3)-(3x-5)(2x+11) b,B=(x^2-2)(x^2+x-1)-x(x^3+x^2-3x-2) Bài 2:Tìm x biết: a,6x(5x+3)+3x(1-10x)=7 b,(3x-3)(5-21x)+(7x+4)(9x-5)=44 c,(x+1)(x+2)(x+5)-x^2(x+8)=27 d,(2x-1)(3-x)+(x-2)(x+3)=(1-x)(x+2) Bài 3 Tính a,(2x+3)^3 b,(x-3y)^3 c.(x+4)(x^2-4x+16) d,(1/3x+2y)(1/9x^2-2/3xy+4y) e,(x-3y)(x2+3xy+9y^2)
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
Tìm x:
a, 3x (4x -3) - 2x (5-6x) = 0
b, 5 (2x-3) + 4x (x-2) + 2x (3-2x) = 0
c, 3x (2-x) + 2x (x-1) = 5x (x+3)
d, 3x (x+1) - 5x (3-x) + 6(x2 + 2x + 3) = 0
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
d) \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
Mà \(14x^2+18>0\)nên pt vô nghiệm