= 9 / 2

\(A=\dfrac{4x+2\sqrt{x}+2}{2\sqrt{x}+1}=\dfrac{2\sqrt{x}\left(2\sqrt{x}+1\right)+2}{2\sqrt{x}+1}=2\sqrt{x}+\dfrac{2}{2\sqrt{x}+1}\)

\(=2\sqrt{x}+1+\dfrac{2}{2\sqrt{x}+1}-1\ge2\sqrt{\left(2\sqrt{x}+1\right)\cdot\dfrac{2}{2\sqrt{x}+1}}-1=2\sqrt{2}-1\)

=> A \(\ge2\sqrt{2}-1\)

Dấu "=" xảy ra <=> \(2\sqrt{x}+1=\dfrac{2}{2\sqrt{x}+1}\)

<=> \(\left(2\sqrt{x}+1\right)^2=2\) <=> \(\left[{}\begin{matrix}2\sqrt{x}+1=2\\2\sqrt{x}+1=-2\left(loại\right)\end{matrix}\right.\)

<=> \(\sqrt{x}=\dfrac{1}{2}\) <=> \(x=\dfrac{1}{4}\)(tm)

Vậy minA = \(2\sqrt{2}-1\) khi x = 1/4

Câu 6:

a: =12x^2+4x-3x-1-5x^2+15x-x^2+7x-12

=6x^2+23x-13

b: =5x^2+5x-2x-2-3x^3+3x^2+9x-2x(x^2-9x+20)

=-3x^3+8x^2+14x-2-2x^3+18x^2-40x

=-5x^3+26x^2-26x-2

3: 

a: \(\Leftrightarrow x+1-6\sqrt{x+1}-9=0\)

=>\(\left(\sqrt{x+1}-3\right)=0\)

=>x+1=9

=>x=8

b: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\left(\sqrt{\dfrac{1}{2}x+1}+3\right)}}=10\)

=>\(\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}-\dfrac{21}{4}}=10\)

=>\(\dfrac{1}{2}x-\dfrac{21}{4}-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}=100\)

=>\(\dfrac{7}{4}\cdot\sqrt{\dfrac{1}{2}x+1}=\dfrac{1}{2}x-\dfrac{21}{4}-100=\dfrac{1}{2}x-\dfrac{421}{4}\)

=>\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{2}{7}x-\dfrac{421}{7}\)

=>1/2x+1=(2/7x-421/7)^2

=>1/2x+1=4/49x^2-1684/49x+177241/49

=>\(x\simeq249,77;x\simeq177,36\)

Giúp chỗ nào >?

giup j zay bn

Tren duong thang xx' lay diem O .tren cung mot nua mat phang bo la duong thang xx' ve 3 tia Oy,Ot,Oz sao cho goc x'oy =40 độ, xOt =97 độ  xoz=54 độ 

chung minh ot la phan giac cua xoy

a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có

góc B chung

=>ΔABC đồng dạng với ΔHBA

=>AB/HB=AC/HA

=>AB*HA=HB*AC

b: AH=căn 5^2-3^2=4cm

BI là phân giác

=>HI/HB=IA/AB

=>HI/3=IA/5=(HI+IA)/(3+5)=0,5

=>HI=1,5cm; IA=1,5cm

\(=\left(\dfrac{1}{49}-\dfrac{1}{9}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)=0\)

1:

AC=căn 5^2-3^2=4cm

BH=AB^2/BC=1,8cm

CH=5-1,8=3,2cm

AH=3*4/5=2,4cm

2:

ΔCBA vuông tại B có tan 40=BC/BA

=>BC/10=tan40

=>BC=8,39(m)

ΔCBD vuông tại B có tan D=BC/BD

=>BD=8,39/tan35=11,98(m)

CTHH của Y có dạng : \(H_xS_yO_z\)

\(x:y:z=\dfrac{\%H}{1}:\dfrac{\%S}{32}:\dfrac{\%O}{16}=\dfrac{2.44}{1}:\dfrac{39.02}{32}:\dfrac{58.54}{16}\\ \\ =2.44:1.22:3.66=2:1:3\)

CT đơn giản nhất là : \(H_2SO_3\)

Bài 17:

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

2) \(x^2-36=\left(x-6\right)\left(x+6\right)\)

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

4) \(25-y^2=\left(5-y\right)\left(5+y\right)\)

5) \(9x^2-1=\left(3x-1\right)\left(3x+1\right)\)

6) \(\dfrac{1}{25}-4x^2=\left(\dfrac{1}{5}-2x\right)\left(\dfrac{1}{5}+2x\right)\)

7) \(9x^2-y^2=\left(3x-y\right)\left(3x+y\right)\)

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

Bài 18:

1) \(\left(x-5\right)\left(x+5\right)=x^2-25\)

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

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

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

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

6) \(-\left(5x-3\right)\left(3+5x\right)=\left(3-5x\right)\left(3+5x\right)=9-25x^2\)

7) \(-\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)=-\left(9x^2-\dfrac{4}{25}\right)=\dfrac{4}{25}-9x^2\)

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

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

2)\(\left(x+3\right)^2\)

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

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

5)\(\left(x-1\right)^2\)

6)\(\left(x-4\right)^2\)

7)\(\left(6-x\right)^2\)

8)\(\left(2x-3y\right)^2\)

9)\(\left(3x-1\right)^2\)