Tìm x biết
a, ( 3x + 2) (2x - 1) - 6x(x - 2) = 3
b, (5x + 1) (6x - 2) = 30x ( x + 1/2)
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( 4x- 1) - 2x(6x- 3 )=30 b) 2x(3-2x) + 2x(2x-1)=15
c) (5x-2)(4x-1) + (10x +3)(2x - 1)=1 d) (x+2) (x+2)- (x -3)(x+1) = 9
e) (4x+1)(6x-3) = 7 + (3x – 2)(8x + 9) g) (10x+2)(4x- 1)- (8x -3)(5x+2) =14
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
a)x4+3/5 - 6x-2/7 = 5x+4/3 +3
b) x-3/x-2 + x-2/x-4 = 3.1/5
c)3/1-4x = 2/4x+1 - 8+6x/16x^2-1
d) x+1/x - x+5/x-2 = 1/x^2 - 2x
bài 2:
a)Tìm m để phương trình 3x+m = x.4 nhận x=-2 là nghiệm
b)Tìm m để phương trình (2x+1)(9x+2k)-5(x+2)=40 có nghiệm x=2
c)Tìm m để phương trình 2mx-3=4x có nghiệm
d)Tìm m để phương trình mx=2-x vô nghiệm
e)Tìm a và b để phương trình a(2x=3)=x+b có nghiệm , cô nghiệm, vô số nghiệm
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
Bài 2 : Tìm x , biết
a) ( 3x -1 ) (2x+7) - ( x +1) (6x-5 ) = 16
b) ( 10x +9 )x - ( 5x -1 ) (2x+3 )= 8
c) ( 3x - 5 ) ( 7- 5x ) + ( 5x +2 )( 3x-2 ) -2 = 0
d) x(x + 1) ( x+6 ) - x3 = 5x
tìm x biết
a) (6x-3) (2x+4) + (4x-1) (5-3x) = -21
b) 6x (3x+5) - 2x (9x-2) + (17-x) (x-1) + x (x-18) =0
c) (15-2x) (4x+1) - (13-4x) (2x-3) - (x-1) (x+2) + x2=52
d) (8x-3) (3x+2) - (4x+7) (x+4) = (2x+1) (5x-1) - 33
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21
<=> 12x2 + 24x - 6x - 12 + 20x - 12x2 - 5 + 3x = -21
<=> 41x = -21 + 12 + 5
<=> 41x = -4
<=> x = -4/41
1.Tính
a, 5x^3yz . (-7x^2y^3)
b, 6x(x-5) -x(6x+3)
c, (x-9)(x^2-2x-1)
2.Cho A (x)=10-2x+4x^3-5x^2
B(x)=-10x^3-5x+6x^2-20
Tính A(x)+B(x); A(x)-B(x)
3.Tìm nghiệm
a,M(x)= 5x+20
b,N(x)=100x^2-49
c,P(x)=3x-15
b. 6x(x - 5) - x(6x + 3)
= x(6x - 30) - x(6x + 3)
= x(6x - 30 - 6x - 3)
= x(-33)
= -33x
1.Tính
a, 5x^3yz . (-7x^2y^3)
b, 6x(x-5) -x(6x+3)
c, (x-9)(x^2-2x-1)
2.Cho A (x)=10-2x+4x^3-5x^2
B(x)=-10x^3-5x+6x^2-20
Tính A(x)+B(x); A(x)-B(x)
3.Tìm nghiệm
a,M(x)= 5x+20
b,N(x)=100x^2-49
c,P(x)=3x-15
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
1.Tính
a, 5x^3yz . (-7x^2y^3)
b, 6x(x-5) -x(6x+3)
c, (x-9)(x^2-2x-1)
2.Cho A (x)=10-2x+4x^3-5x^2
B(x)=-10x^3-5x+6x^2-20
Tính A(x)+B(x); A(x)-B(x)
3.Tìm nghiệm
a,M(x)= 5x+20
b,N(x)=100x^2-49
c,P(x)=3x-15
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
Bài 3
1.(x-1)(x+2)+5x-5=0
2.(3x+5)(x-3)-6x-10=0
3.(x-2)(2x+3)-7x2+14x=0
4.(x+1)(x-3)-15+5x=0
5.5(2x-1)(x+3)+5x-10x2=0
Bài4
1.3x-6+(x+1)(x-2)=0
2.4x2+6x-(2x+3)(3x-x)=0
3.5x-10-(2-x)(4+x)=0
4.10-10x+(x-1)(x-3)=0
5.20x2+30x-2(x-5)(2x+3)=0
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)