Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
-x2-6x+11
=-(x2+6x)+11
=-[(x2+2.x.3+9)-9]+11
=-[[(x+3)2-9]]+11
= -(x+3)2 + 9+11
= - (x+3)2 + 20
Ở dấu = thứ 3 bạn lm sai r nha.
Khi tách 1 hạng tử ở trong ngoặc ra ngoài ngoặc mà trc ngoặc có dấu trừ thì phải đổi dấu, do đó -9 trở thành 9
1,(2x + 3 ) \(^{^{ }2}\)=\(\left(2x\right)^2+2.2x.3+3^2\)
=\(4x^2+12x+9\)
2, ( 3x + 2y )\(^2=\left(3x\right)^2+2.3x.2y+\left(2y\right)^2\)
=\(9x^2+12xy+4y^2\)
3,(3a -1 )\(^2=\left(3a\right)^2-2.3a.1+1^2\)
\(=9a^2-6a+1\)
4, (a - 2 )\(^2=a^2-2.a.2+2^2\)
=\(a^2-4a+4\)
5, ( 1 - 5a )\(^2=1^2-2.1.5a+\left(5a\right)^2\)
=\(1-10a+25a\)
6, ( x - 4 )\(^3=x^3-3x^24+3x4^2-4^3\)
=\(x^3-12x^2+48x-64\)
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
\(a.\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\\\Leftrightarrow 9x^2+12x+4-9x^2+12x-4=5x+38\\ \Leftrightarrow24x-5x=38\\ \Leftrightarrow19x=38\\\Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\\Leftrightarrow 3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-3x^2-12x+9x-3x=-12+9-9\\ \Leftrightarrow-6x=-12\\\Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(c.\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x-2\right)\\ \Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x+22\\ \Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x+22\\\Leftrightarrow x^3-x^3-3x^2-2x^2+5x^2+3x-x-10x+11x=1+22\\ \Leftrightarrow3x=23\\\Leftrightarrow x=\frac{23}{3}\)
Vậy nghiệm của phương trình trên là \(\frac{23}{3}\)
\(d.\left(x+3\right)^2-\left(x-3\right)^2=6x+18\\ \Leftrightarrow x^2+6x+9-x^2+6x-9=6x+18\\ \Leftrightarrow12x-6x=18\\ \Leftrightarrow6x=18\\ \Leftrightarrow x=3\)
Vậy nghiệm của phương trình trên là \(3\)
\(e.\left(x+1\right)\left(x^2-x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\\\Leftrightarrow x^3+1-2x=x\left(x^2-1\right)\\\Leftrightarrow x^3+1-2x=x^3-x\\ \Leftrightarrow x^3-x^3-2x+x=-1\\ \Leftrightarrow-x=-1\\ \Leftrightarrow x=1\)
Vậy nghiệm của phương trình trên là \(1\)
\(f.\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\\\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\\ \Leftrightarrow x^3-x^3-6x^2+9x^2-3x^2+12x-3x=8+1+1\\ \Leftrightarrow9x=10\\ \Leftrightarrow x=\frac{10}{9}\)
Vậy nghiệm của phương trình trên là \(\frac{10}{9}\)
Bài 1. ( 27x3 - 8) : ( 9x2 + 6x + 4)
= [ ( 3x)3 - 23] : ( 9x2 + 6x + 4)
= ( 3x - 2)( 9x2 + 6x + 4) : ( 9x2 + 6x + 4)
= 3x - 2
Bài 2. A = ( 3x - 5)( 2x + 11) - ( 2x + 3)( 3x + 7)
A = 6x2 + 33x - 10x - 55 - ( 6x2 + 23x + 27)
A = - 28
KL......
B = (2x+3)( 4x2- 6x + 9) -2( 4x3-1)
B = 8x3 + 27 - 8x3 + 2
B = 29
KL......
C= (x - 1)3-(x + 1)3+6( x + 1)( x - 1 )
C = ( x - 1)( x2 - 2x + 1 + 6x + 6) - ( x + 1)3
C = ( x - 1)( x2 + 4x + 7 ) - ( x + 1)3
C = x3 + 4x2 + 7x - x2 - 4x - 7 - x3 - 3x2 - 3x - 1
C = - 8
KL.........
\(B=\frac{x^2+10x+20}{x^2+6x+9}=\frac{(x^2+6x+9)+4(x+3)-1}{x^2+6x+9}\)
\(=1+\frac{4(x+3)}{x^2+6x+9}-\frac{1}{x^2+6x+9}=1+\frac{4(x+3)}{(x+3)^2}-\frac{1}{(x+3)^2}\)
\(=1+\frac{4}{(x+3)}-\frac{1}{(x+3)^2}\)
Đặt \(\frac{1}{x+3}=a\Rightarrow B=1+4a-a^2=5-(a^2-4a+4)\)
\(=5-(a-2)^2\leq 5\)
Vậy \(B_{\max}=5\Leftrightarrow a=2\Leftrightarrow x=-\frac{5}{2}\)
\(C=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}\)
Có: \(3x^2+9x+7=3(x^2+3x+\frac{9}{4})+\frac{1}{4}=3(x+\frac{3}{2})^2+\frac{1}{4}\geq \frac{1}{4}\)
\(\Rightarrow \frac{10}{3x^2+9x+7}\leq \frac{10}{\frac{1}{4}}=40\)
\(\Rightarrow C\leq 41\)
Vậy \(C_{\max}=41\Leftrightarrow x=\frac{-3}{2}\)
b) \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x^3+3x^2+3x^2+9x=0\)
\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=0\end{cases}}}\)
Vậy \(x\in\left\{-3;0\right\}\)
a) \(2x\left(x-2\right)+x^2=4\)
\(\Leftrightarrow2x\left(x-2\right)+x^2-4=0\)
\(\Leftrightarrow2x\left(x-2\right)+\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)
Vậy \(x\in\left\{\frac{-2}{3};2\right\}\)
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
nhanh mình đang cần gấp