a, A=-x^2 -11x+7 =-(x^2 +11x -7) = -(x^2 +11x +121/4) +149/4 >= 149/4

A(max)=149/4 tại x=-11/2

b: Ta có: \(-5x^2+7x+8\)

\(=-5\left(x^2-\dfrac{7}{5}x-\dfrac{8}{5}\right)\)

\(=-5\left(x^2-2\cdot x\cdot\dfrac{7}{10}+\dfrac{49}{100}-\dfrac{209}{100}\right)\)

\(=-5\left(x-\dfrac{7}{10}\right)^2+\dfrac{209}{20}\le\dfrac{209}{20}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{7}{10}\)

a: Ta có: \(-x^2-11x+7\)

\(=-\left(x^2+11x-7\right)\)

\(=-\left(x^2+2\cdot x\cdot\dfrac{11}{2}+\dfrac{121}{4}-\dfrac{149}{4}\right)\)

\(=-\left(x+\dfrac{11}{2}\right)^2+\dfrac{149}{4}\le\dfrac{149}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{11}{2}\)

c: Ta có: \(-7x^2+8x+15\)

\(=-7\left(x^2-\dfrac{8}{7}x-\dfrac{15}{7}\right)\)

\(=-7\left(x^2-2\cdot x\cdot\dfrac{4}{7}+\dfrac{16}{49}-\dfrac{121}{49}\right)\)

\(=-7\left(x-\dfrac{4}{7}\right)^2+\dfrac{121}{7}\le\dfrac{121}{7}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{4}{7}\)

bạn lớp 8 đó hả có chơi ff ko

trl giúp mình vs ạ :)))))

thay x = 1/2, ta có:

B = 27.(1/2)^3 + 28.(1/2)^2 + 11.1/2 + 2

B = 143/8

$[27\cdot(-67)+33\cdot(-27)]:(-30)$

$=[(-27)\cdot67+33\cdot(-27)]:(-30)$

$=[-27\cdot(67+33)]:(-30)$

$=(-27\cdot100):(-30)$

$=-2700:(-30)$

$=2700:30$

$=90$

ko bit moi hoc lop 3

Giải Phương Trình Sau (Nhớ ghi cách làm nha mình k đúng cho)

Giải Phương Trình Sau (Nhớ ghi cách làm nha mình ĐÁNH DẤU ĐÚNG  cho)

b) 3x⁴-3x³+9x³-9x²-24x²+24x-48x+48

=3x³(x-1)+9x²(x-1)-24x(x-1)-48(x-1)

=(x-1)(3x³+9x²-24x-48)

=3(x-1)(x³+3x²-8x-16)

a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

b, \(đk:x\ge2\)

Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0

 \(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)

\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)

\(\Leftrightarrow x^3-11x^2+35x-25\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\)  (*)

Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)

Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5

c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)

\(\Leftrightarrow4x^3+x>0\)

Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))

\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)

\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....

d) Đk: \(x\ge\dfrac{3}{4}\)

Áp dụng bđt cosi:

 \(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)

 \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)

\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)

\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)

Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)

Dấu = xảy ra khi x=1 (tm)

`a)\sqrtx+\sqrt{2-x}=(3x^2-2x+3)/(x^2+1)`

`đk:0<=x<=2`

`pt<=>sqrtx-1+\sqrt{2-x}-1=(3x^2-2x+3)/(x^2+1)-2`

`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x^2-2x+1)/(x^2+1)`

`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x-1)^2/(x^2+1)`

`<=>(x-1)((x-1)/(x^2+1)+1/(sqrt{2-x}+1)-1/(sqrtx+1))=0`

`<=>x-1=0<=>x=1`

Vậy `S={1}`

e, \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)

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

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

d, \(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

a) \(\dfrac{4y^2}{11x^4}:\left(-\dfrac{8y}{33x^2}\right)\)

\(=\dfrac{4y^2}{11x^4}.\left(-\dfrac{33x^2}{8y}\right)\)

\(=-\dfrac{4y^2.33x^2}{11x^4.8y}\)

\(=-\dfrac{3y}{2x^2}\)

b) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\dfrac{x+4}{2\left(x-2\right)}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}\)

\(=\dfrac{x+2}{6}\).

\(\dfrac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\dfrac{x+2}{6}\)