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16 tháng 6 2023
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
L
2
22 tháng 9 2020
a) \(\left(x+2\right)\left(x+3\right)-\left(x+1\right)\left(x+7\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-8x-7=6\)
\(\Leftrightarrow-3x=7\)
\(\Leftrightarrow x=-\frac{7}{3}\)
b) \(\left(8x-3\right)\left(3x+2\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(\Leftrightarrow\left(8x-3\right)\left(9x^2+12x+4\right)-4x^2-23x-28=10x^2+3x-1-33\)
\(\Leftrightarrow72x^3+69x^2-4x-12-14x^2-26x+6=0\)
\(\Leftrightarrow72x^3+55x^2-30x-6=0\)
Nghiệm vô tỉ: \(x_1=-1,078...\) ; \(x_2=0,476...\) ; \(x_3=-0,162...\)
22 tháng 9 2020
a) (x + 2)(x + 3) - (x + 1)(x + 7) = 6
=> x(x + 3) + 2(x + 3) - x(x + 7) - 1(x + 7) = 6
=> x2 + 3x + 2x + 6 - x2 - 7x - x - 7 = 6
=> x2 + 5x + 6 - x2 - 7x - x - 7 = 6
=> (x2 - x2) + (5x - 7x - x) + (6 - 7) = 6
=> -3x - 1 = 6
=> -3x = 7
=> x = -7/3
b) (8x - 3)(3x + 2)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
=> (8x - 3)(9x2 + 12x + 4) - [4x(x + 4) + 7(x + 4)] = 2x(5x - 1) + 1(5x - 1) - 33
=> 8x(9x2 + 12x + 4) - 3(9x2 + 12x + 4) - (4x2 + 16x + 7x + 28) = 10x2 - 2x + 5x - 1 - 33
=> 72x3 + 96x2 + 32x - 27x2 - 36x - 12 - 4x2 - 16x - 7x - 28 - 10x2 + 2x - 5x + 1 + 33 = 0
=> 72x3 + (96x2 - 27x2 - 10x2 - 4x2) + (32x - 36x - 16x - 7x + 2x - 5x) + (-12 - 28 + 1 + 33) = 0
=> 72x3 + 55x2 - 30x - 6 = 0
=> x vô nghiệm
NT
3
17 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33`
`\Leftrightarrow 8x(3x+2) -3(3x+2) - 4x(x+4) + 7(x+4) = 2x(5x-1) + 5x-1 - 33`
`\Leftrightarrow 24x^2 + 16x - 9x - 6 - 4x^2 - 16x - 7x - 28 = 10x^2 - 2x + 5x - 1 - 33`
`\Leftrightarrow 20x^2 -16x - 34 = 10x^2 + 3x - 34`
`\Leftrightarrow 20x^2 - 16x - 34 - 10x^2 - 3x + 34 = 0`
`\Leftrightarrow 10x^2 - 19x = 0`
`\Leftrightarrow x(10x - 19)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)
Vậy, `x={0; 19/10}.`
PT
1
NV
31 tháng 8 2015
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
2 tháng 1 2016
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
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2019
a/ \(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\frac{22}{5}\)
b/ \(\Leftrightarrow-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c/ \(\Leftrightarrow24x^2+7x-6-4x^2-9x+28=10x^2+3x-1-33\)
\(\Leftrightarrow10x^2-5x+56=0\)
Phương trình vô nghiệm (chắc do bạn ghi sai đề)
HQ
18 tháng 9 2019
a/ ⇔2x2−5x−12+x2−7x+10=3x2−17x+20⇔2x2−5x−12+x2−7x+10=3x2−17x+20
⇔5x=22⇔5x=22
⇒x=225⇒x=225
b/ ⇔−5x2−2x+16+4x2−4x−8+2x2−8=0⇔−5x2−2x+16+4x2−4x−8+2x2−8=0
⇔x2−6x=0⇔x2−6x=0
⇔x(x−6)=0⇒[x=0x=6⇔x(x−6)=0⇒[x=0x=6
c/ ⇔24x2+7x−6−4x2−9x+28=10x2+3x−1−33⇔24x2+7x−6−4x2−9x+28=10x2+3x−1−33
⇔10x2−5x+56=0⇔10x2−5x+56=0
Phương trình vô nghiệm (chắc do bạn ghi sai đề)
DH
11 tháng 7 2017
a, \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2-4\right)=0\)
\(\Rightarrow16-5x^2-2x+4x^2-4x-8+2x^2-8=0\)
\(\Rightarrow x^2-6x=0\Rightarrow x.\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy.............
b, \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(\Rightarrow24x^2+16x-9x-6-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1-33\)
\(\Rightarrow24x^2+7x-6-4x^2-23x-28-10x^2-3x=-1-33\)
\(\Rightarrow10x^2-19x=-1-33+28+6\)
\(\Rightarrow x.\left(10x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)
Vậy..........
Chúc bạn học tốt!!!
6 tháng 6 2017
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 0
<=> x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 0
<=> x2 + 3x + 2x + 6 - x2 - 5x + 2x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> (2x + 3)(x - 4) + (x - 5)(x - 2) - (3x - 5)(x - 4) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - (3x2 - 12x - 5x + 20) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - 3x2 + 12x + 5x - 20 = 0
<=> 5x = 12 - 10 + 20
<=> 5x = 22
<=> x = 22/5
c) (8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
<=> 8x + 16 - 5x2 - 10x + (4x - 8)(x + 1) + 2(x2 - 4) = 0
<=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0
<=> x2 - 6x = 0
<=> x(x - 6) = 0
<=> x = 0 hay x - 6 = 0
I<=> x = 6
d) (8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
<=> 24x2 + 16x - 9x - 6 - (4x2 + 16x + 7x + 28) = 10x2 - 2x + 5x - 1 - 33
<=> 24x2 + 16x - 9x - 6 - 4x2 - 16x - 7x - 28 - 10x2 + 2x - 5x + 1 + 33 = 0
<=> 10x2 - 19x = 0
<=> x(10x - 19) = 0
<=> x = 0 hay 10x - 19 = 0
I <=> 10x = 19
I <=> x = 19/10
\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)+\left(2x+1\right)\left(1-5x\right)=-33\)
\(pt\Leftrightarrow3x\left(8x-3\right)+2\left(8x-3\right)-\left(x\left(4x+7\right)+4\left(4x+7\right)\right)+\left(2x+1\right)-5x\left(2x+1\right)+33=0\)
\(\Leftrightarrow24x^2-9x+16x-6-\left(4x^2+7x+16x+28\right)+2x+1-10x^2-5x+33=0\)
\(\Leftrightarrow24x^2-9x+16x-6-4x^2-7x-16x-28+2x+1-10x^2-5x+33=0\)
\(\Leftrightarrow10x^2-19x=0\Leftrightarrow x\left(10x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)