em mới học lớp 6 thôi anh ơi

em mới vô lớp 6 thui mà òa

cai nay la hag dag thuc phan tih ra la dk

pt<=>căn((x-1/2)^2+75/4)+căn(2(x-1/2)^2+3(x+2)^2)+căn((x-1/2)^2+3(2x+3/2)^2)>=3*căn3(x+2)

dấu = xãy ra khi x=1/2

A=\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+1\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\) ( với \(x^4-8x^2-9=x^4-9x^2+x^2-9=x^2\left(x^2-9\right)+\left(x^2-9\right)=\left(x^2-9\right)\left(x^2+1\right)=\left(x-3\right)\left(x+3\right)\left(x^2+1\right)\)  

A= \(\frac{13-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}-\frac{2}{x-3}=0\) \(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\) \(\Leftrightarrow\left(10x-30\right)\left(x-3\right)+6-2\left(x+3\right)=0\Leftrightarrow-x^2+11x-30=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=5\end{array}\right.\)

Mày nhìn cái chóa j

ĐK: \(x\ne-3,3,-2\)

Ta có: \(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{x^2+5x+6}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-9x^2+x^2-9}-\frac{3x+6}{x^2+3x+2x+6}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^2.\left(x^2-9\right)+\left(x^2-9\right)}-\frac{3x+6}{x.\left(x+3\right)+2.\left(x+3\right)}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6.\left(x^2+1\right)}{\left(x^2+1\right).\left(x^2-9\right)}-\frac{3.\left(x+2\right)}{\left(x+2\right).\left(x+3\right)}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6}{x^2-9}-\frac{3}{x+3}-\frac{2}{x-3}=0\)

=>\(\left(\frac{13-x}{x+3}-\frac{3}{x+3}\right)+\left(\frac{6}{x^2-9}-\frac{2}{x-3}\right)=0\)

=>\(\frac{13-x-3}{x+3}+\left[\frac{6}{x^2-9}-\frac{2.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\right]=0\)

=>\(\frac{10-x}{x+3}+\left[\frac{6}{x^2-9}-\frac{2x+6}{x^2-9}\right]=0\)

=>\(\frac{10-x}{x+3}+\frac{6-2x-6}{x^2-9}=0\)

=>\(\frac{\left(10-x\right).\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}+\frac{-2x}{x^2-9}=0\)

=>\(\frac{13x-x^2-30}{x^2-9}-\frac{2x}{x^2-9}=0\)

=>\(\frac{13x-x^2-30-2x}{x^2-9}=0\)

=>\(\frac{11x-x^2-30}{x^2-9}=0\)

Vì \(x\ne-3,3=>x^2\ne0\)

=>11x-x²-30=0

=>6x-30-x²+5x=0

=>6.(x-5)-x.(x-5)=0

=>(6-x).(x-5)=0

=>6-x=0=>x=6

hoặc x-5=0=>x=5

Vậy tập nghiệm của phương trình S=6; 5

Em ước gì được ên lớp 8 để giúp anh  Hoàng Phúc

Nobita Kun nó lớp 7 đó cha ko làm thì thôi ai bắt vô cmt

\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)

\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)

\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)

Nên \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

đex ~ vừa thấy trên face lướt qua luôn

\(\sqrt{x^2-3x+2}-\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)

\(\Leftrightarrow\left(\sqrt{x^2-3x+2}-\sqrt{x-2}\right)-\left(\sqrt{x^2+2x-3}+\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)-\left(x-2\right)}{\sqrt{x^2-3x+2}+\sqrt{x-2}}-\dfrac{\left(x^2+2x-3\right)-\left(x+3\right)}{\sqrt{x^2+2x-3}-\sqrt{x+3}}=0\)

\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x-2}}-\dfrac{\left(x-2\right)\left(x+3\right)}{\sqrt{\left(x+3\right)\left(x-1\right)}-\sqrt{x+3}}=0\)

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{x-2}{\sqrt{x-2}\left(\sqrt{x-1}+1\right)}-\dfrac{x+3}{\sqrt{x+3}\left(\sqrt{x-1}-1\right)}\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\right]=0\)

Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô n_o

(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))

=> x - 2 = 0

<=> x = 2 (nhận)

\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)

\(\Leftrightarrow\dfrac{\left(4x+1\right)-\left(3x-2\right)}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)

\(\Leftrightarrow\dfrac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}\right)\left(x+3\right)=0\)

TH1:

x + 3 = 0

<=> x = - 3 (loại)

TH2:

\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)

\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\)

\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)+\left(\sqrt{3x-2}-2\right)=0\)

\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)

\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{3\left(x-2\right)}{\sqrt{3x-2}+2}=0\)

\(\Leftrightarrow\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}\right)\left(x-2\right)=0\)

Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô n_o

=> x - 2 = 0

<=> x = 2 (nhận)

~ ~ ~

Vậy x = 2

\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

\(\Leftrightarrow\sqrt{2\left(x^2+4x+3\right)}-\left[\left(2x+2\right)-\sqrt{x^2-1}\right]=0\)

\(\Leftrightarrow\sqrt{2\left(x+3\right)\left(x+1\right)}-\dfrac{\left(4x^2+8x+4\right)-\left(x^2-1\right)}{\sqrt{x^2-1}+2x+2}=0\)

\(\Leftrightarrow\sqrt{2\left(x+3\right)\left(x+1\right)}-\dfrac{\left(x+1\right)\left(3x+5\right)}{\sqrt{\left(x-1\right)\left(x+1\right)}+2\left(x+1\right)}=0\)

\(\Leftrightarrow\sqrt{x+1}\left[2\sqrt{x+3}-\dfrac{\sqrt{x+1}\left(3x+5\right)}{\sqrt{x+1}\left(\sqrt{x-1}+2\sqrt{x+1}\right)}\right]=0\)

\(\Leftrightarrow\sqrt{x+1}\left[2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\right]=0\)

TH1

x + 1 = 0

<=> x = - 1 (loại)

TH2

\(2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}=0\)

mà \(2\sqrt{x+3}=\dfrac{4x+12}{2\sqrt{x+3}}>\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\forall x\ge1\)

=> VT > 0

=> vô n_o

~ ~ ~

Vậy pt vô n_o