(2x+3)^4=81
Vậy x =
Tính giá trị biểu thức:
a) (-15) + (-35) + x biết x = -31
Thay x = -31 vào biểu thức đã cho ta có:
(-15) + (-35) + (-31) = - 81
Vậy GTBT đã cho là – 81 tại x = - 31.
b) (-2) + 17 + x biết x = -28
c) x + 25 + (-47) biết x = -4
d) 25 + x + (-13) biết x = 8
Mn làm cho em câu b ; c ; d với ạ!
b) Thay x=-28
\(\left(-2\right)+17+\left(-28\right)=-13\)
c) Thay x=-4
\(\left(-4\right)+25+\left(-47\right)=-26\)
d) Thay x=8
\(25+8+\left(-13\right)=20\)
Đa thức P(x) = 2x^4 + 3x^2 − x^3 − 3x^4 − x^2 − 2x + 1 sau khi được thu gọn và sắp xếp theo bậc giảm dần của biến là:
A. P(x) = x^4 − x^3 + 2x^2 − 2x + 1
B.P(x) = −x^4 − x^3 + 3x^2 − 2x + 1
C. P(x) = −x^4 − x^3 + 2x^2 − 2x + 1
D. P(x) = x^4 − x^3 − 2x^2 − 2x + 1
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
a. ( 5 - 2x ) ( 5x + 2x ) + 2x ( x + 3 ) = 4 - 2x² b. ( 3x - 2 )( -2x) + 5x² = -x( x - 3) c. 7 - ( 3 + 2x ) (2x - 3 ) = ( x + 4 )²
a: Sửa đề: (5-2x)(5+2x)+2x(x+3)=4-2x^2
=>25-4x^2+2x^2+6x=4-2x^2
=>6x+25=4
=>6x=-21
=>x=-7/2
b: (3x-2)(-2x)+5x^2=-x(x-3)
=>-6x^2+4x+5x^2=-x^2+3x
=>4x=3x
=>x=0
c: =>7-(4x^2-9)=x^2+8x+16
=>7-4x^2+9-x^2-8x-16=0
=>-5x^2-8x=0
=>5x^2+8x=0
=>x(5x+8)=0
=>x=0 hoặc x=-8/5
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
mấy bn check hộ mk, nếu sai hãy chỉ ra giùm và sử lại nhé,
A=|x-2|+|2x-3|+|3x-4| ap dung BĐT|x|+|y|>=|x+y| |x-2|+|3x-4|=|x-2|+|4-3x|>=|x-2+4-3x|=|-2x-2| =)A= |x-2|+|4-3x|+|2x-3|>=|-2x+2|+|2x-3|>=|-2x+2+2x-3| =|-1|=1 dau = xay ra khi (x-2)(4-3x)>=0 va (-2x+2)(2x-3)>=0 =)4/3
+) Lỗi nhỏ: Sai ở chỗ: \(\left|x-2+4-3x\right|=\left|-2x-2\right|\)
+) Lỗi lớn: Dấu bằng xảy ra: \(\hept{\begin{cases}\left(x-2\right)\left(4-3x\right)\ge0\\\left(-2x+2\right)\left(2x-3\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{3}{2}\le x\le1\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le1\)( làm tắt )
Nhưng mà thử vào chọn x= 1=> A = 3 > 1. Nên bài này sai.
Làm lại nhé!
A = | x - 2 | + | 2 x - 3 | + | 3 x - 4 |
= | x - 2 | + | 2 x - 3 | + 3 | x - 4/3 |
= | x -2 | + | x - 4/3 | + | 2x -3 | +2 | x - 4/3 |
= ( | 2 - x | + | x - 4/3 | ) + ( | 3 - 2x | + | 2x - 8/3 | )
\(\ge\)| 2 -x + x - 4/3 | + | 3 - 2x + 2x -8/3 |
= 2/3 + 1/3 = 1
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(2-x\right)\left(x-\frac{4}{3}\right)\ge0\\\left(3-2x\right)\left(2x-\frac{8}{3}\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{4}{3}\le x\le\frac{3}{2}\end{cases}}\Leftrightarrow\frac{4}{3}\le x\le\frac{3}{2}\)
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: \(x=-2\)
Tính (rút gọn )
1, 2x(3x-1)-(2x+1)(x-3)
2, 3(x^2-2x)-(4x+2)(x-1)
3, 3x(x-5)-(x-2)^2 -(2x+3)(2x-3)
4, (2x-3)^2+(2x-1) (x+4)
1) `2x(3x-1)-(2x+1)(x-3)`
`=6x^2-2x-2x^2+6x-x+3`
`=4x^2+3x+3`
2) `3(x^2-3x)-(4x+2)(x-1)`
`=3x^2-9x-4x^2+4x-2x+2`
`=-x^2-7x+2`
3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`
`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`
`=3x^2-15x-x^2+4x-4-4x^2+9`
`=-2x^2-11x+5`
4) `(2x-3)^2+(2x-1)(x+4)`
`=4x^2-12x+9+2x^2+8x-x-4`
`=6x^2-5x+5`
1) 2x – (3 – 5x) = 4( x +3)
2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)
3) 5x - 4(6-x)(x + 3) = (4-2x)(3-2x) + 2
4) (x - 1)3 - (3x + 2)(-12) = (x2 + 1)(x - 2) - x2
5) (3x -1)2 - (x +3)(2x-1) = 7(x + 1)(x -2) -3x
mn giúp mình vs
1) 2x – (3 – 5x) = 4( x +3)
<=>2x-3+5x=4x+12
<=>2x-3+5x-4x-12=0
<=>3x-15=0
<=>x=5
2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)
<=>10x-15-20x+28=19-2x-22
<=>10x-15-20x+28-19+2x+22=0
<=>-8x+16=0
<=>x=2
tham khảo
1) 2x – (3 – 5x) = 4( x +3)
<=>2x-3+5x=4x+12
<=>2x-3+5x-4x-12=0
<=>3x-15=0
<=>x=5
2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)
<=>10x-15-20x+28=19-2x-22
<=>10x-15-20x+28-19+2x+22=0
<=>-8x+16=0
<=>x=2
tìm x biết
a,5x(x-4)3(x+2)(x-4)=2x(x+1)
b,4x(x+2)-x(8x-5)=10
c,(x+3)(2x-5)=2x(x+4)
d,(3x-2)(x+5)-3x(x+4)=5
e,x(x-3)+2x(x+1)=3(x mũ2-4)
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12