VT>=0 suy ra 4x>=0

suy ra x>=0

..................................................................................................

Do : VP ≥ 0

=> VT ≥ 0

=> 4x ≥ 0

=> x ≥ 0

nên Phương trình trên có dạng :

x + 2 + x + 9 + x + 2011 = 4x

<=> 3x + 2022 = 4x

<=> x = 2022 ( thỏa mãn )

KL....

nghiệm nguyên hay k nguyên bn

\(\left(2-x\right)\left(2x-1\right)+\left(4x^2-4x+1\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(2x-1\right)+\left(2x-1\right)^2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2-x+2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-1\end{array}\right.\)

Vậy phương trình có tập nghiệm \(\left\{-1;\frac{1}{2}\right\}\)

(2-x)(2x-1)+(4x^2-4x+1)=0

 Ta có:  (2x-1)(2-x)+(2x-1)^2=0

                (2x-1)(2-x+2x-1)=0

Sau đó bn tự lam nha tại vì mk làm bằng phone

sáng sớm lang thang lật lại mấy trang gặp bài này, xin trình bày vài cách:

Đk:\(x\ge2\) \(\left(DK\forall PP\right)\)

C1 \(pt\Leftrightarrow x^3-3x\left(x+2\right)-2\sqrt{\left(x+2\right)^3}=0\)

Đặt \(t=\sqrt{x+2}\) ra pt đăng cấp bậc 3...

c2:\(pt\Leftrightarrow\left(\sqrt{\left(x+2\right)^3}+1\right)^2=\left(3\left(x+1\right)\right)^2\)

c3:\(pt\Leftrightarrow\left(\sqrt{\left(x+2\right)^3}-3x-2\right)\left(3x+\sqrt{\left(x+2\right)^3+4}\right)=0\)

C4:Chia 2 vế x³ dc:

\(1-\frac{3}{x}\pm2\sqrt{\left(\frac{1}{x}+\frac{2}{x^2}\right)}-\frac{6}{x^2}=0\)

đặt \(\sqrt{\left(\frac{1}{x}+\frac{2}{x^2}\right)}=t\) dc \(1\pm3t^2+2t^3=0\)

Ngoài ra còn có thể liên hợp ,.....

a, \(\Leftrightarrow\left(x+1+x-2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)+\left(x-2\right)^2\right]-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-\left(2x-1\right)^2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-4x^2+4x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(-3x^2+3x+6\right)=0\)

\(\Leftrightarrow-3\left(2x-1\right)\left(x^2-x-2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)\left(x-2\right)=0\)

=>x=1/2 hoặc x=-1 hoặc x=2

Vậy pt có tập nghiệm là S={1/2;-1;2}

b, \(x^4=24x+32\Leftrightarrow x^4-24x-32=0\)

\(\Leftrightarrow x^4-2x^3-4x^2+2x^3-4x^2-8x+8x^2-16x-32=0\)

\(\Leftrightarrow x^2\left(x^2-2x-4\right)+2x\left(x^2-2x-4\right)+8\left(x^2-2x-4\right)=0\)

\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)

\(\Leftrightarrow x^2-2x-4=0\) (vì x^2+2x+8 > 0)

\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x-1=\pm\sqrt{5}\Leftrightarrow x=1\pm\sqrt{5}\)

Vậy...

c, \(\left(x-6\right)^4+\left(x-8\right)^4=16\)

Đặt x-6=t => x-8=t-2

Ta có: \(t^4+\left(t-2\right)^4=16\Leftrightarrow t^4+t^4-8t^3+24t^2-32t+16=16\)

\(\Leftrightarrow2t^4-8t^3+24t^2-32t=0\Leftrightarrow t^4-4t^3+12t^2-16t=0\)

\(\Leftrightarrow t^4-2t^3-2t^3+4t^2+8t^2-16t=0\)

\(\Leftrightarrow t^3\left(t-2\right)-2t^2\left(t-2\right)+8t\left(t-2\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t^3-2t^2+8t\right)=0\Leftrightarrow\left(t-2\right)t\left(t^2-2t+8\right)=0\)

Mà t^2-2t+8=(t-1)^2+7 > 0

\(\Rightarrow\orbr{\begin{cases}t-2=0\\t=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-6-2=0\\x-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)

Vậy...

Từ pt (1) \(\Rightarrow x=8+\left|y-5\right|\ge8\Rightarrow x+1>0\)

- Nếu \(y\ge5\Rightarrow3\left|y+3\right|\ge24>21\Rightarrow\) vô nghiệm

- Nếu \(-5\le y\le5\) hệ trở thành:

\(\left\{{}\begin{matrix}x-\left(5-y\right)=8\\x+1+3\left(y+5\right)=21\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=13\\x+3y=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=17\\y=-4\end{matrix}\right.\)

- Nếu \(y< -5\) hệ trở thành:

\(\left\{{}\begin{matrix}x-\left(5-y\right)=8\\x+1+3\left(-y-5\right)=21\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=13\\x-3y=35\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{37}{2}\\y=\dfrac{-11}{2}\end{matrix}\right.\)

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\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)

\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)

\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)

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

\(\Leftrightarrow-x^2-3=3x-3\)

\(\Leftrightarrow-x^2-3x=0\)

\(\Leftrightarrow-x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\) 

Vậy x = 0 

\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)

\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)

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

\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow-6x+3=8\)

\(\Leftrightarrow x=-\frac{5}{6}\)

Vậy ... 

sorry, em mới học lớp 6 thui ạ

em mời hok lớp 7 thôi ạ