Những hằng đẳng thức đáng nhớ - Hỏi đáp

a) Ta có: \(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

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

b) Ta có: \(x^5+x-1\)

\(=x^5+x^2-x^2+x-1\)

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

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

\(=\left(x^2-x+1\right)\left[x^2\left(x+1\right)-1\right]\)

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

c) Ta có: \(a^3+b^3+c^3-3abc\)

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

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[a^2+2ab+b^2-ca-cb+c^2-3ab\right]\)

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

\(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

Ta có:

\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

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

\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)

\(=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\) (đpcm)

=$-3(x^2+25/3$ $x$)

=$-3(x^2+2.25/6$ $x$ +$625/36$ -$625/36$

=$-3$ $(x+23/6)^2$+$625/12$ <$<$ $625/12$ $AX$

=$MaxB$ = $625/12$ => $x$ = $-25/6$

Học tốt nhé em

\(B=-3x^2-25x\)

\(=-3\left(x^2+\frac{25}{3}x\right)\)

\(=-3\left(x^2+2.\frac{25}{6}x+\frac{625}{36}-\frac{625}{36}\right)\)

\(=-3\left(x+\frac{25}{6}\right)^2+\frac{625}{12}\le\frac{625}{12}\forall x\)

=> \(MaxB=\frac{625}{12}\Leftrightarrow x=-\frac{25}{6}\)

a) \(\left(x+y\right)^2+\left(x-y\right)^2\)

=\(\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)

=\(x^2+2xy+y^2+x^2-2xy+y^2\)

\(2x^2+2y^2=2\left(x^2+y^2\right)\)

b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

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

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

\(=2x^2\)

c) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)

\(=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2\)

= \(\left(x-y+z-z+y\right)^2=x^2\)

\(a,\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3x^2\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=\left(x-2y\right)^2\)

a,( 2x-3)²-(x+5)² = 0

=> (2x-3-x-5)(2x-3+x+5)=0

=> (x-8)(3x+2)=0

=> \(\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\3x=-2\Rightarrow x=\dfrac{-2}{3}\end{matrix}\right.\)

vậy x=8 hoăc x=\(\dfrac{-2}{3}\)

b, (x³-x²) - 4x²+8x-4 =0

=> x²(x-1)-(4x²-8x+4)=0

=> x²(x-1)-4(x²-2x+1)=0

=> x²(x-1)-4(x-1)²=0

=> (x-1)[x²-4(x-1)]=0

=> (x-1)(x²-4x+4)=0

=> (x-1)(x-2)²=0

=> \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

vậy x=1; x=2

Tìm x ak

a,(2x-3)²-(x+5)²=0

->(2x-3+x+5)(2x-3-x-5)=0

->(3x+2)(x-8)=0

=>3x+2=0 hoặc x-8=0

->x=-2/3 hoặc x=8

b,b, (x³-x²) - 4x²+8x-4 =0

=>x²(x-1)-4(x²-2x+1)=0

=>x²(x-1)-4(x-1)²=0

=>(x-1)(x²-4x+4)=0

=>x-1=0 hoặc (x-2)²=0

=>x=1 hoặc x=2

a, 8x ( x- 2017) - 2x + 4034 = 0

⇔ 8x(x - 2017) - 2(x - 2017) = 0

⇔ ( x - 2017 ) (8x - 2) = 0

⇔ \(\left[{}\begin{matrix}x-2017=0\\8x-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=2017\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy...

b, \(\frac{x}{2}+\frac{x^2}{8}=0\)

⇔ \(\frac{4x}{8}+\frac{x^2}{8}=0\)

⇔ 4x + x² = 0

⇔ x(4 + x) = 0

⇔ \(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

c, 4 - x = 2(x-4)^2

⇔ 4 - x = 2 ( x² - 8x + 16)

⇔ 4 - x = 2x² - 16x + 32

⇔ 2x² - 15x + 28 = 0

⇔ 2x² - 8x - 7x + 28 = 0

⇔ 2x(x - 4) - 7(x - 4) = 0

⇔ (2x - 7) (x - 4) = 0

⇔ \(\left[{}\begin{matrix}x=\frac{7}{2}\\x=4\end{matrix}\right.\)

Vậy..

d, ( x^2 + 1) ( x-2) + 2x = 4

⇔ x³ + x - 2x² - 2 + 2x = 4

⇔ x³ - 2x² + 3x - 6 = 0

⇔ x²( x - 2) + 3( x - 2) = 0

⇔ (x² + 3)(x - 2) = 0

⇔ x - 2 = 0

⇔ x = 2

Vậy....

a) 3x(x + 2)

= 3x.x + 3x.2

= 3x² + 6x

b) 4x(x - 5)

= 4x.x + 4x.(-5)

= 4x² - 20x

c) x(x + 3)

= x.x + x.3

= x² + 3x

d) x(x + 5)

= x.x + x.5

= x² + 5x

e) (x + 5)(x + 1)

= x(x + 1) + 5(x + 1)

= x.x + x.1 + 5.x + 5.1

= x² + x + 5x + 5

= x² + 6x + 5

f) (x + 2)(x + 3)

= x(x + 3) + 2(x + 3)

= x.x + x.3 + 2.x + 2.3

= x² + 3x + 2x + 6

= x² + 5x + 6

g) (x + 3)(x + 4)

= x(x + 4) + 3(x + 4)

= x.x + x.4 + 3.x + 3.4

= x² + 4x + 3x + 12

= x² + 7x + 12

h) (x + 1)(x - 2)

= x(x - 2) + 1(x - 2)

= x.x + x.(-2) + 1.x + 1.(-2)

= x² - 2x + x - 2

= x² - x - 2

\(2x^3-12x^2+18x=0\)

\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

vậy.,........

c/m biểu thức sau là dương:

a) \(x^2-8x+20\) = \(x^2-8x+16+4\)=\(\left(x-4\right)^2+4\ge4>0\)

Vậy biểu thức trên là dương.

b) \(4x^2-12x+11\)\(=4x^2-12x+9+2\)= \(\left(2x-3\right)^2+2\ge2>0\)

Vậy biểu thức trên dương.

c) \(x^2-x+1\)\(=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

Vậy biểu thức trên dương.

d) \(x^2-2x+y^2+4y+6\)

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

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\) mà 1>0

=> \(x^2-2x+y^2+4y+6>0\)

Vậy biểu thức trên dương.

Nếu đề là chứng minh \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) thì nó đúng.

(Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2abc}=0\))

Còn nếu như comment của bạn bên dưới bình luận của bạn Nhi thì đề bạn sai nha=))

Chứng minh gì

a) \(A= 2x^2- 3x +1\)

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

\(=2\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{1}{16}\right)\)

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

Vậy A_min = \(-\dfrac{1}{8}\) khi \(x=\dfrac{3}{4}\)

b) \(B= 4x^2 +7x + 13\)

\(=\left(2x\right)^2+2\cdot2x\cdot\dfrac{7}{4}+\dfrac{49}{16}+\dfrac{159}{16}\)

\(=\left(2x+\dfrac{7}{4}\right)^2+\dfrac{159}{16}\ge\dfrac{159}{16}\)

Vậy B_min = \(\dfrac{159}{16}\) khi \(x=-\dfrac{7}{8}\)

c) \(C= 5-8x+x^2\)

\(=x^2-2\cdot x\cdot4+16+9\)

\(=\left(x-4\right)^2+9\ge9\)

Vậy C_min = 9 khi x = 4

d) \(D = (x-1)(x+2)(x+3)(x+6)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\)

Vậy D_min = - 36 khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Do chiều rộng là x + 5 (m), chiều dài là 3x - 10 (m) và diện tích hình chữ nhật là 300 m² nên:

(x + 5).(3x - 10) = 300

3x² - 10x + 15x - 50 - 300 = 0

3x² + 5x - 350 = 0

3x² - 30x + 35x - 350 = 0

3x(x - 10) + 35(x - 10) = 0

(x - 10)(3x + 35) = 0

x - 10 = 0 hoặc 3x + 35 = 0

* x - 10 = 0

x = 0 + 10

x = 10

* 3x + 35 = 0

3x = 0 - 35

3x = -35

x = -35/3

Với x = 10, ta có:

Chiều dài hình chữ nhật là:

3.10 - 10 = 20 (m)

Chiều rộng hình chữ nhật là:

10 + 5 = 15 (m)

Chu vi hình chữ nhật là:

(20 + 15).2 = 70 (m)

Với x = -35/3, ta có:

Chiều dài hình chữ nhật là:

3.(-35/3) - 10 = -45 (vô lý vì độ dài cạnh không thể âm)

Vậy chu vi hình chữ nhật là 70 m

x² + 2x –24=x² + 2x+1-5²

=(x+1)²-5²

=(x+1-5)(x+1+5)

=(x-4)(x+6)

CHÚC BẠN HỌC TỐT

$x^2+2x-24$

$=x^2+2x+1-25$

$=(x+1)^2-5^2$

$=(x+1+5)(x+1-5)$

$=(x+6)(x-4)$