Biết rằng \left(x^{4}-4\right)\left(-2x^{3}+5x-3\right) = -2x^{7}+5x^{5}-3x^{4}+8x^{3}-20x+12(x4−4)(−2x3+5x−3)=−2x7+5x5−3x4+8x3−20x+12.
Kết quả phép nhân \left(-x^{4}+4\right)\left(-2x^{3}+5x-3\right)(−x4+4)(−2x3+5x−3) là ?
Mọi người ơi, giúp mình với mình đang cần gấp
thank mọi người
1 thực hiện phép nhân
1)\(-3\left(-x+3\right)\)
2)\(-5x^3\left(-3x+5\right)\)
3)\(-2x\left(-2x-6\right)\)
4)\(-3x^3\left(-2x-12\right)\)
5)\(-10x\left(-5x+2\right)\)
6)\(-2x^2\left(-3x^3+4x^2-5\right)\)
7)\(-x^2\left(-2x^3-x+3\right)\)
8)\(-2^3\left(-2-6\right)\)
9)\(\left(-x-3\right)\left(x+2\right)\)
10)\(\left(2x-3\right)\left(5-x\right)\)
11)\(\left(-x+6\right)\left(-x-2\right)\)
12)\(\left(3x-1\right)\left(-3-2x\right)\)
13)\(\left(-5x-3\right)\left(2-x\right)\)
14) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
15) \(\left(x-3\right)\left(1-2x-5y\right)\)
16)\(\left(x-2\right)\left(x^2+4\right)\left(x+2\right)\)
17)\(\left(-3+1\right)\left(9x^2+1\right)\left(-3x-1\right)\)
18)\(-\left(2x+1\right)\left(2-1\right)\left(x+1\right)\)
19)\(\left(2x^2-3x+5\right)\left(x^2-8x+2\right)\)
20)\(\left(3x^2y-6xy+9x\right)\left(-xy\right)\)
1. -3(-x+3)
= 3x - 6
2. -5x3 (-3x + 5)
= 15x4 - 25x3
3. -2x (-2x - 6)
= 4x2 + 12x
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)
\(\Leftrightarrow42x-39=4\)
\(\Leftrightarrow42x=4+39\)
\(\Leftrightarrow42x=43\)
\(\Rightarrow x=\dfrac{43}{42}\)
Tick cho mk nha
\(\frac{-3x.\left(5x+3\right)}{1+3x}>=0\)\(\frac{-2x^2+5x-3}{-x.\left(3x+7\right)}>0\)\(\frac{1}{x-2}-\frac{4}{x^2-4}< \frac{1}{3}\)\(x^2-20x+51>0\)\(\left(x-3\right).\left(2x+1\right)\left(1-5x\right)< 0\)\(\left(x-2\right)\left(x+3\right)=< 0\)
Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
Bài 3 : Xét dấu biểu thức sau :
1 , \(f\left(x\right)=\frac{x-7}{4x^2-19x+12}\)
2 , \(f\left(x\right)=\frac{11x+3}{-x^2+5x-7}\)
3 , \(f\left(x\right)=\frac{3x-2}{x^3-3x^2+2}\)
4 , \(f\left(x\right)=\frac{x^2+4x-12}{\sqrt{6}x^2+3x+\sqrt{2}}\)
5 , \(f\left(x\right)=\frac{x^2-3x-2}{-x^2+x-1}\)
6 , \(f\left(x\right)=\frac{x^3-5x+4}{x^4-4x^3+8x-5}\)
7 , \(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left(-2x^2+x-1\right)}{\left(2x-5\right)\left(x^2+3x-10\right)}\)
8 , \(f\left(x\right)=\left(-x^2+x-1\right)\left(6x^2-5x+1\right)\)
9 , \(f\left(x\right)=\frac{x^2-x-2}{-x^2+3x+4}\)
10 , \(f\left(x\right)=\left(x^2-5x+4\right)\left(2-5x+2x^2\right)\)
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
5.
\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)
\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Rút gọn các phân thức :
a) \(\dfrac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
b) \(\dfrac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}\)
c) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
d) \(\dfrac{5x^2-10xy}{2\left(2y-x\right)^3}\)
e) \(\dfrac{32x-8x^2+2x^3}{x^3+64}\)
f) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
g) \(\dfrac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
h) \(\dfrac{5x^3+5x}{x^4-1}\)
i) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
Tìm x, biết:
1) \(\left|4x\right|=3x+12\) 7) \(\left|5x\right|-3x-2=0\)
2) \(\left|2x+4\right|=2x-5\) 8) \(x-5x+\left|-2x\right|-3=0\)
3)\(\left|x+3\right|=3x-1\) 9) \(\left|3-x\right|+x^2-\left(4+x\right)x=0\)
4) \(\left|x-4\right|+3x=5\)
5)\(\left|x-5\right|=3x\)
6) \(\left|x+2\right|=2x-10\)
Làm tiếp nè :
2) / 2x + 4/ = 2x - 5
Do : / 2x + 4 / ≥ 0 ∀x
⇒ 2x - 5 ≥ 0
⇔ x ≥ \(\dfrac{5}{2}\)
Bình phương hai vế của phương trình , ta có :
( 2x + 4)2 = ( 2x - 5)2
⇔ ( 2x + 4)2 - ( 2x - 5)2 = 0
⇔ ( 2x + 4 - 2x + 5)( 2x + 4 + 2x - 5) = 0
⇔ 9( 4x - 1) = 0
⇔ x = \(\dfrac{1}{4}\) ( KTM)
Vậy , phương trình vô nghiệm .
3) / x + 3/ = 3x - 1
Do : / x + 3 / ≥ 0 ∀x
⇒ 3x - 1 ≥ 0
⇔ x ≥ \(\dfrac{1}{3}\)
Bình phương hai vế của phương trình , ta có :
( x + 3)2 = ( 3x - 1)2
⇔ ( x + 3)2 - ( 3x - 1)2 = 0
⇔ ( x + 3 - 3x + 1)( x + 3 + 3x - 1) = 0
⇔ ( 4 - 2x)( 4x + 2) = 0
⇔ x = 2 (TM) hoặc x = \(\dfrac{-1}{2}\) ( KTM)
KL......
4) / x - 4/ + 3x = 5
⇔ / x - 4/ = 5 - 3x
Do : / x - 4/ ≥ 0 ∀x
⇒ 5 - 3x ≥ 0
⇔ x ≤ \(\dfrac{-5}{3}\)
Bình phương cả hai vế của phương trình , ta có :
( x - 4)2 = ( 5 - 3x)2
⇔ ( x - 4)2 - ( 5 - 3x)2 = 0
⇔ ( x - 4 - 5 + 3x)( x - 4 + 5 - 3x) = 0
⇔ ( 4x - 9)( 1 - 2x) = 0
⇔ x = \(\dfrac{9}{4}\) ( KTM) hoặc x = \(\dfrac{1}{2}\) ( KTM)
KL......
Làm tương tự với các phần khác nha
1)\(\left|4x\right|=3x+12\)
\(\Leftrightarrow4.\left|x\right|=3x+12\\ \Leftrightarrow4.\left|x\right|-3x=12\)
\(TH1:4x-3x=12\left(x\ge0\right)\\\Leftrightarrow x=12\left(TM\right) \)
\(TH2:4.\left(-x\right)-3x=12\left(x< 0\right)\\ \Leftrightarrow-7x=12\\ \Leftrightarrow x=-\dfrac{12}{7}\left(TM\right)\)
Vậy tập nghiệm của PT: \(S=\left\{12;-\dfrac{12}{7}\right\}\)