1-x-2x^2

= 1-x-2x.2x

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

= 1 - 5x

Để 1-x-2x^2 mang giá trị lớn nhất thì x phài là số âm.

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

\(=-2\left(x^2+2\times x\times\frac{1}{4}+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2-\frac{1}{2}\right)\)

\(=-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)

\(\left(x+\frac{1}{4}\right)^2\ge0\)

\(\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\ge-\frac{9}{16}\)

\(-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\le\frac{9}{8}\)

Vậy Max A = \(\frac{9}{8}\) khi x = \(-\frac{1}{4}\)

b: Ta có: \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)

\(\Leftrightarrow50x=-100\)

hay x=-2

1: BC=căn 7,5^2+10^2=12,5cm

AH=7,5*10/12,5=6cm

góc ADH=góc AEH=góc DAE=90 độ

=>ADHE là hcn

=>DE=AH=6cm

2: ΔAHB vuông tại H có HD là đường cao

nên AD*AB=AH^2

ΔAHC vuông tại H có HElà đường cao

nên AE*AC=AH^2

=>AD*AB=AE*AC

3: ΔAHB vuông tại H có HD là đường cao

nên DH^2=AD*DB

4: AD=AH^2/AB=6^2/7,5=36/7,5=4,8cm

AE=AH^2/AC=6^2/10=3,6cm

S AED=1/2*3,6*4,8=8,64cm²

g, PT \(\Leftrightarrow\dfrac{x+24}{1996}+1+\dfrac{x+25}{1995}+1+\dfrac{x+26}{1994}+1+\dfrac{x+27}{1993}+1+\dfrac{x+2036}{4}-4=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{1996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy ...

1: \(\left(2x+1\right)^2=\left(2x\right)^2+2\cdot2x\cdot1+1^2=4x^2+4x+1\)

2: \(\left(3x+\frac13y\right)^2=\left(3x\right)^2+2\cdot3x\cdot\frac13y+\left(\frac13y\right)^2=9x^2+2xy+\frac19y^2\)

3: \(\left(xy-\frac12y^2\right)^2\)

\(=\left(xy\right)^2-2\cdot xy\cdot\frac12y^2+\left(\frac12y^2\right)^2\)

\(=x^2y^2-xy^3+\frac14y^4\)

4: \(\left(-\frac32x+2\right)^2=\left(\frac32x-2\right)^2\)

\(=\left(\frac32x\right)^2-2\cdot\frac32x\cdot2+2^2=\frac94x^2-6x+4\)

5: \(\left(5x-\frac14y\right)\left(5x+\frac14y\right)=\left(5x\right)^2-\left(\frac14y\right)^2=25x^2-\frac{1}{16}y^2\)

6: \(\left(x^2+\frac23x\right)\left(\frac23x-x^2\right)=\left(\frac23x\right)^2-\left(x^2\right)^2=\frac49x^2-x^4\)

7: \(\left(2x+y\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

\(=8x^3+12x^2y+6xy^2+y^3\)

8: \(\left(1-\frac13y\right)^3=1^3-3\cdot1^2\cdot\frac13y+3\cdot1\cdot\left(\frac13y\right)^2-\left(\frac13y\right)^3\)

\(=1-y+\frac13y^2-\frac{1}{27}y^3\)

9: \(\left(x+4\right)\left(x^2-4x+16\right)\)

\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)

\(=x^3+4^3=x^3+64\)

10: \(\left(\frac45-5x\right)\left(\frac{16}{25}+4x+25x^2\right)=\left(\frac45-5x\right)\left\lbrack\left(\frac45\right)^2+\frac45\cdot5x+\left(5x\right)^2\right\rbrack\)

\(=\left(\frac45\right)^3-\left(5x\right)^3=\frac{64}{125}-125x^3\)

11: \(\left(1-x-y\right)^2=\left(x+y-1\right)^2=x^2+y^2+1+2xy-2x-2y\)

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

\(=x^2-y^2-z^2-2yz\)

\(a,f\left(x\right)+g\left(x\right)\\ =10x^5-5x^5-8x^4+2x^4+6x^3-4x^3-4x^2+6x^2+2x-8x+1+10+3x^6+2x^6\\ =5x^6+5x^5-6x^4+2x^3+2x^2-6x+11\\ f\left(x\right)-g\left(x\right)\\ =3x^6-2x^6+10x^5+5x^5-8x^4-2x^4+6x^3+4x^3-4x^2-6x^2+2x+8x+1-10\\ =x^6+15x^5-10x^4+10x^3-10x^2+10x-9\)

\(b,f\left(x\right) +g \left(x\right)=3x^4+2x^4+15x^3-15x^3+7x^2-7x^2+3x-3x-\dfrac{1}{2}+\dfrac{1}{2}=5x^4\\ f\left(x\right)-g\left(x\right)=3x^4-2x^4+15x^3+15x^3+7x^2+7x^2+3x+3x-\dfrac{1}{2}-\dfrac{1}{2}\\ =x^4+30x^3+14x^2+6x-1\)

Em cần giúp câu nào hả em? Em nên chụp 1-2 ý cho 1 lần hỏi nhá, như thế mọi người sẽ dễ dàng giúp em hơn

13

a, \(3x-4=-x+8\)

\(< =>3x+x=8+4\)

\(< =>4x=12\)

\(< =>x=\frac{12}{4}=3\)

b, \(\frac{2x+1}{6}+\frac{x-7}{12}=10\)

\(< =>\frac{2\left(2x+1\right)}{12}+\frac{x-7}{12}=\frac{120}{12}\)

\(< =>4x+2+x-7=120\)

\(< =>5x=120+5=125\)

\(< =>x=\frac{125}{5}=\frac{5^3}{5}=5^2=25\)

a) \(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+10x+4}{x}\left(x\ne0;x\ne-2\right)\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{\left(x+2\right)-x^2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{-x^2+x+2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)}{x}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-10x-4}{x}\)

\(Q=\dfrac{-x^3-2x^2-6x}{x}\)

\(Q=\dfrac{x\left(-x^2-2x-6\right)}{x}\)

\(Q=-x^2-2x-6\)

b) Ta có:

\(Q=-x^2-2x-6\)

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

\(Q=-\left[\left(x^2+2x+1\right)+5\right]\)

\(Q=-\left(x+1\right)^2-5\)

Mà: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow Q=-\left(x+1\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi:

\(x+1=0\Rightarrow x=-1\)

Vậy: \(Q_{max}=-5\Leftrightarrow x=-1\)

Câu 5: B

Câu 6: B

Cậu giải đc bài 2 hoặc 3 phần tự luận ko ạ