Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)

HĐT không được phép quên \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)

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\(\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=14\end{matrix}\right.\)\(\Rightarrow\left(ab+bc+ac\right)=-7\)

\(\left\{{}\begin{matrix}a+b+c=0\\\left(ab+bc+ac\right)=-7\end{matrix}\right.\)\(\Rightarrow\left(ab+bc+ac\right)^2=\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2=7^2\)

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\right]\)

\(a^4+b^4+c^4=14^2-2.7^2=7^2\left(4-2\right)=2.7^2\)

a/ \(\left(x+\dfrac{1}{x}\right)^2=x^2+2+\dfrac{1}{x^2}=2+7=9\)

\(\Rightarrow x+\dfrac{1}{x}=\pm3\)

Với \(x+\dfrac{1}{x}=3\) thì ta có:

\(\left(x+\dfrac{1}{x}\right)^3=x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=x^3+\dfrac{1}{x^3}+9\)

\(\Rightarrow x^3+\dfrac{1}{x^3}=27-9=18\)

Ta có: \(\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=x+\dfrac{1}{x}+x^5+\dfrac{1}{x^5}=3+x^5+\dfrac{1}{x^5}\)

\(\Rightarrow x^5+\dfrac{1}{x^5}=7.18-3=123\)

Tương tự cho trường hợp còn lại ta được ĐPCm

b/ Thay \(y=0,75x-2,5\) vào A rồi rút gọn ta được A sau đó làm như bình thường là được

cách khác

đề không bắt tìm x^5+1/x^5

\(\left(x^2+\dfrac{1}{x^2}\right)=7\in N\Rightarrow\Rightarrow\left(x+\dfrac{1}{x}\right)^n\in N\Rightarrow\left(x^n+\dfrac{1}{x^n}\right)\in N\forall n\in N\)=> dpcm

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2ac+2bc\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\forall a;b;c\)

Vậy \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\) đpcm

VT = (a+b+c)³-a³-b³-c³

= \([\left(a+b\right)+c]^3\)- a³-b³-c³

= (a+b)³+c³ +3ab(a+b)+3c(a+b)(a+b+c)-a³-b³-c³

⁼3(a+b) \([ab+c\left(a+b+c\right)]\)

= 3(a+b) \([ab+ac+bc+c^2]\)

= 3(a+b)(b+c)(c+a)

\(\Rightarrow\)VT=VP= 3(a+b)(b+c)(c+a)

Giải:

Ta có: \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)

\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)

\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ca\right)\)

\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\) (Đpcm)

a) 2x².(5x³-4x²y-7xy +1) =10x⁵-8x⁴y-14x³y+2x² b) (5x -2y)(x² -xy +1) =5x³-5x²y+5x-2x²y+2xy²-2y =5x³-7x²y+2xy²+5x-2y c) (\(\dfrac{1}{2}\)x -1)(2x -3) =x²-\(\dfrac{3}{2}\)x-2x+3 =x²-\(\dfrac{7}{2}\)x+3 d) (x +3y)² =x²+6xy+9y² e) (3x -2y)² =9x²-12xy+4y² g) (\(\dfrac{1}{4}\)x - 3y)(\(\dfrac{1}{4}\)x +3y) =\(\dfrac{1}{16}\)x²-9y² f) (2x +3)³ =8x³+36x²+54x+27 h) (3 -2y)³ =27-54y+36y²-8y³

a, \(x^2+6x+9=\left(x+3\right)^2\)

b, \(4x^2-4x+1=\left(2x-1\right)^2\)

c, \(\dfrac{1}{8}-x^3=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{64}+\dfrac{1}{2}x+x^2\right)\)

b, \(64x^3-\dfrac{1}{27}=\left(4x-\dfrac{1}{3}\right)\left(16x^2+\dfrac{4}{3}x+\dfrac{1}{9}\right)\)

a, \(x^2+6x+9=\left(x+3\right)^2\)

b, \(4x^2-4x+1=\left(2x-1\right)^2\)

c, \(\dfrac{1}{8}-x^3=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{4}+\dfrac{1}{2}x+x^2\right)\)

b, \(64x^3-\dfrac{1}{27}=\left(4x-\dfrac{1}{3}\right)\left(16x^2+\dfrac{4}{3}x+\dfrac{1}{9}\right)\)

a, \(x^2+6x+9=\left(x+3\right)^2\)

b, \(4x^2-4x+1=\left(2x-1\right)^2\)

c, \(\dfrac{1}{8}-x^3=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{4}+\dfrac{1}{2}x+x^2\right)\)

d, \(64x^3-\dfrac{1}{27}=\left(4x-\dfrac{1}{3}\right)\left(16x^2+\dfrac{4}{3}x+\dfrac{1}{9}\right)\)

Q không có giá trị nhỏ nhất.

Vì \(2^2-6x\)\(\Leftrightarrow4-6x\) mà x là hằng số => 4 - 6x có rất nhiều giá trị vô hạn.

Bạn xem lại đề nhé.

Có phải đề của bạn là \(x^2-6x\)

Mình sửa đề:

\(Q=x^2-6x=x^2-3x-3x+9-9\)

\(Q=\left(x^2-3x\right)-\left(3x-9\right)-9\)

\(Q=x.\left(x-3\right)-3.\left(x-3\right)-9\)

\(Q=\left(x-3\right)^2-9\)

Với mọi giá trị của \(x\in R\) ta có:

\(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-9\ge-9\)

Hay \(Q\ge-9\) với mọi giá trị của \(x\in R\).

Để \(Q=-9\) thì \(\left(x-3\right)^2-9=-9\)

\(\Rightarrow\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\)

Vậy GTNN của biểu thức là \(-9\) đạt được khi và chỉ khi \(x=3\).

Chúc bạn học tốt!!!

Đặt \(A=4x-x^2+3\)

\(=-x^2+4x+3=-\left(x^2-4x-3\right)\)

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left[\left(x-2\right)^2-7\right]\)

\(=-\left(x-2\right)^2+7\)

Ta có: \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2+7\le7\)

Dấu " = " khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy \(MAX_A=7\) khi x = 2

Đặt \(A=-3x^2+2x-1\)

\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)

\(=-3\left(x^2-2.x.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)

\(=-3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{2}{9}\right]\)

\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{2}{3}\)

Ta có: \(-3\left(x-\dfrac{1}{3}\right)^2\le0\)

\(\Rightarrow A=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{2}{3}\le\dfrac{-2}{3}\)

Dấu " = " xảy ra khi \(-3\left(x-\dfrac{1}{3}\right)^2=0\Leftrightarrow x=\dfrac{1}{3}\)

Vậy \(MAX_A=\dfrac{-2}{3}\) khi \(x=\dfrac{1}{3}\)

Gtnn và gtln là j vậy ?

b)\(B=1^2-2^2+3^2-4^2+...-2016^2+2017^2\)

\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+...+\left(2015^2-2016^2\right)+2017^2\)

\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2015-2016\right)\left(2015+2016\right)+2017^2\)

\(=-1\cdot\left(1+2\right)+\left(-1\right)\cdot\left(3+4\right)+...+\left(-1\right)\cdot\left(2015+2016\right)+2017^2\)

\(=-1\cdot\left(1+2+...+2015+2016\right)+2017^2\)

\(=-1\cdot\dfrac{2016\cdot\left(2016+1\right)}{2}+2017^2\)

\(=-2033136+4068289=2035153\)

c)\(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(=2^{64}-1-2^{64}=-1\)