\(P\left(x\right)=5x^2+x+2=5\left(x^2+\dfrac{1}{5}x\right)+2\\ =5\left(x^2+2.x.\dfrac{1}{10}+\left(\dfrac{1}{10}\right)^2\right)-5.\left(\dfrac{1}{10}\right)^2+2\\ =5\left(x+\dfrac{1}{10}\right)^2+\dfrac{39}{20}\)

Nhận xét: \(\left(x+\dfrac{1}{10}\right)^2\ge0\forall x\inℝ\\ \Rightarrow5\left(x+\dfrac{1}{10}\right)^2\ge0\\ \Rightarrow P\left(x\right)=5\left(x+\dfrac{1}{10}\right)^2+\dfrac{39}{20}\ge\dfrac{39}{20}\)

\(Min_{P\left(x\right)}=\dfrac{39}{20}\) tại \(\left(x+\dfrac{1}{10}\right)^2=0\Leftrightarrow x+\dfrac{1}{10}=0\Leftrightarrow x=-\dfrac{1}{10}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\left(1\right)\\ \dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\dfrac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) => \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Sửa đề: Oz là tia đối của tia Oy

Ot là phân giác của góc xOy

=>\(\widehat{yOt}=\dfrac{\widehat{xOy}}{2}=\dfrac{110^0}{2}=55^0\)

Ta có: \(\widehat{yOt}+\widehat{tOz}=180^0\)(hai góc kề bù)

=>\(\widehat{tOz}+55^0=180^0\)

=>\(\widehat{tOz}=125^0\)

a: ta có: \(BM=MC=\dfrac{BC}{2}\)

\(CN=ND=\dfrac{CD}{2}\)

mà BC=CD

nên BM=MC=CN=ND

Xét ΔABM vuông tại B và ΔBCN vuông tại C có

AB=BC

BM=CN

Do đó: ΔABM=ΔBCN

=>AM=BN

ΔABM=ΔBCN

=>\(\widehat{BMA}=\widehat{CNB}\)

=>\(\widehat{AMB}+\widehat{CBN}=90^0\)

=>AM\(\perp\)BN tại E

chịu

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

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

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

\(=-4x^2+x+1\)

Em cần làm gì với biểu thức này thì nên ghi rõ yêu cầu ra em nhé!

g) (3x + 4)(3x - 4) - (2x + 5)² = (x - 5)² + (2x + 1)² - (x² - 2x) + (x - 1)²

9x² - 16 - 4x² - 20x - 25 = x² - 10x + 25 + 4x² + 4x + 1 - x² + 2x + x² - 2x + 1

5x² - 20x - 41 = 5x² - 6x + 27

5x² - 5x² - 20x + 6x = 27 + 41

-14x = 68

i) -5(x + 3)² + (x - 1)(x + 1) + (2x - 3)² = (5x - 2)² - 5x(5x + 3)

-5(x² + 6x + 9) + x² - 1 + 4x² - 12x + 9 = 25x² - 20x + 4 - 25x² - 15x

-5x² - 30x - 45 + x² - 1 + 4x² - 12x + 9 = -35x + 4

-42x - 37 = -35x + 4

-42x + 35x = 4 + 37

-7x = 41

a) (-x + 5)(x - 2) + (x - 7)(x + 7) = (3x + 1)² - (3x - 2)(3x + 2)

-x² + 2x + 5x - 10 + x² - 49 = 9x² + 6x + 1 - 9x² + 4

7x - 59 = 6x + 5

7x - 6x = 5 + 59

x = 64

b) (5x - 1)(x + 1) - 2(x - 3)² = (x + 2)(3x - 1) - (x + 4)² + (x² - x)

5x² + 5x - x - 1 - 2(x² - 6x + 9) = 3x² - x + 6x - 2 - x² - 8x - 16 + x² - x

5x² + 4x - 1 - 2x² + 12x - 18 = 3x² - 4x - 18

3x² + 16x - 19 = 3x² - 4x - 18

3x² + 16x - 3x² + 4x = -18 + 19

20x = 1

a.

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

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

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

b.

Từ câu a:

\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow a^4+b^4+c^4=\dfrac{\left(a^2+b^2+c^2\right)^2}{2}\)

\(\left(x-2\right)^2-\left(x+3\right)^2+\left(x+4\right)\left(x-4\right)=0\\ < =>x^2-4x+4-x^2-6x-9+x^2-16=0\\ < =>x^2-10x-21=0\\ < =>\left(x^2-10x+25\right)-46=0\\ < =>\left(x-5\right)^2=46\\ < =>\left[{}\begin{matrix}x-5=\sqrt{46}\\x-5=-\sqrt{46}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\sqrt{46}+5\\x=5-\sqrt{46}\end{matrix}\right.\)