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1. a) 1012 - 992 = (101 + 99)(101 - 99) = 200 . 2 = 400
b) 98.102 = (100 - 2)(100 + 2) = 1002 - 4 = 10000 - 4 = 9996
c) 772 + 232 + 77.46 = 772 + 232 + 77.23.2 = (23 + 77)2 = 1002 = 10000
d) M = x3 + 9x2 + 27x + 27 = (x + 3)3 = (7 + 3)3 = 103 = 1000
2. a) 2x2 + 3x - 5 = 0
=> 2x2 + 5x - 2x - 5 = 0
=> x(2x + 5) - (2x + 5) = 0
=> (x - 1)(2x + 5) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
b) 2x2 - 11x - 51 = 0
=> 2x2 - 17x + 6x - 51 = 0
=> x(2x - 17) + 3(2x - 17) = 0
=> (x + 3)(2x - 17) = 0
=> \(\orbr{\begin{cases}x+3=0\\2x-17=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=\frac{17}{2}\end{cases}}\)
a) 1012 - 992 = (101-99)(101+99)= 2,200 = 4002
b)98.102 = (100-2)(100+2) = 1002 - 22 =10000 - 4 = 9996
c) 772 + 232 +77.46 = 772 + 232 +2.77.23 = ( 77+23)2 = 1002 =1000
d) Với x=7 => M = 73+ 9.73 + 27.7 + 27 = 10.73 +27.8 = 10.343 + 216 = 3430+216 = 3646
2. a) 2x2 + 3x -5 =0
=> 2(x2 +3/2 x +9/16) -49/8 = 0
=> 2 (x+3/4)2 =49/8
=> (x+3/4)2 =49/16 = (7/4)2 = (-7/4)2
=> x+3/4 = 7/4 hoặc x+3/4 = -7/4
=> x= 1 hoặc x=-5/2
b) 2x2 -11x - 51 =0
=> 2(x2 -11/2x + 121/16) -529/8 = 0
=> (x -11/4)2 = 529/16 = (23/4)2 =(-23/4)2
=> x-11/4=23/4 hoặc x-11/4 = -23/4
=> x=17/2 hoặc x=-3
a) \(\left(x+2\right)^2-\left(x+4\right)^2=0\)
\(\Rightarrow\left(x+2-x-4\right)\left(x+2+x+4\right)=0\)
\(\Rightarrow\left(-2\right)\left(2x+6\right)=0\)
\(\Rightarrow\left(-2\right).2.\left(x+3\right)=0\)
\(\Rightarrow x+3=0\) (vì \(-4\ne0\) )
\(\Rightarrow x=-3\)
Vậy \(x=-3\) (câu này mk có sửa đề ko biết có đúng ko 
!!!)
b) \(\left(x-3\right)^2-9=0\Rightarrow\left(x-3\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=3^2\\\left(x-3\right)^2=\left(-3\right)^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
Vậy \(x=6\) hoặc \(x=0\)
c) \(x^2+6x+9=0\Rightarrow\left(x+3\right)^2=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
Vậy \(x=-3\)
d) \(-x^3+9x^2-27x+27=0\)
\(\Rightarrow-\left(x^3-9x^2+27x-27\right)=0\)
\(\Rightarrow-\left(x-3\right)^3=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
A = x2 +3x+3 min
<=>( x^2 +2x.3/2 + 9/4 ) -9/4 +3
<=> (x+3/2)^2 + 3/4 >= 3/4 ((x+3/2)^2>=0)
dấu "="xảy ra khi x=-3/2
vậy Pmin=3/4 khi x=-3/2
\(a,A=\left(x+5\right)^3\)
\(b,B=\left(x-3\right)^3\)
\(c,C=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}=\left(\frac{x}{2}+\frac{y}{3}\right)^3\)
Mk nghĩ đề bài phần c fải như trên ,cn đâu bn tự thay số vào nha.
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) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
\(x^2+y^2-xy-2x-2y+9=x^2+y^2+2xy-2x-2y+9-3xy\)
\(=\left(x+y\right)^2-2\left(x+y\right)+9-3xy=\left(x+y-2\right)\left(x+y\right)+9-3xy.\)
\(đếnđâytịt\)
b
c, =3 dễ
\(\frac{3x^2-6x+9}{x^2-2x+3}=\frac{3\left(x^2-2x+3\right)}{x^2-2x+3}=3\)
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
b1 A=10000 B=23386
b2 a,x=5
b,x=4 x=2
có lẽ bn nên tự lm thì hơn
a: \(x^2-10x=-25\)
\(\Leftrightarrow x-5=0\)
hay x=5
b: \(x^2-6x+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)