ĐK: \(x^3+4x^2+5x+6\ge0\)

Ta có: \(x^3+4x^2+5x+6=\left(x+3\right)\left(x^2+x+2\right);x^2+2x+5=\left(x+3\right)+\left(x^2+x+2\right)\)

Đặt \(\hept{\begin{cases}\sqrt{x+3}=u\\\sqrt{x^2+x+2}=v\end{cases}}\)

Vậy nên ta có phương trình: \(\)\(u^2+v^2=\frac{5}{2}uv\)

\(\Leftrightarrow2u^2-5uv+2v^2=0\Leftrightarrow\orbr{\begin{cases}u=2v\\u=\frac{1}{2}v\end{cases}}\)

Với u = 2v ta có: \(\sqrt{x+3}=2\sqrt{x^2+x+2}\Leftrightarrow x+3=4x^2+4x+8\)

\(\Leftrightarrow4x^2+3x+5=0\)   (Vô nghiệm)

Với \(u=\frac{1}{2}v\) ta có: \(2\sqrt{x+3}=\sqrt{x^2+x+2}\Leftrightarrow4x+12=x^2+x+2\)

\(\Leftrightarrow x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\left(tmđk\right)\)

Vậy phương trình có nghiệm \(x\in\left\{5;-2\right\}\)

d) \(F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)(1)

Đặt \(x^2+x+1=t\)

\(\Rightarrow\left(1\right)=t\left(t+1\right)-12=t^2+t-12\)

\(=t^2+4t-3t-12\)

\(=t\left(t+4\right)-3\left(t+4\right)=\left(t-3\right)\left(t+4\right)\)(2)

Mà \(x^2+x+1=t\)(ẩn phụ)

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

c) \(E=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)

Đặt \(x^2+7x+10=t\)

\(\Rightarrow\left(1\right)=t\left(t+2\right)-24=t^2+2t-24\)

\(=t^2+6t-4t-24\)

\(=t\left(t+6\right)-4\left(t+6\right)=\left(t-4\right)\left(t+6\right)\)(2)

Mà \(x^2+7x+10=t\)(ẩn phụ)

Nên \(\left(2\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

b)\(D=x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)(1)

Đặt \(x^2+10x=t\)

\(\Rightarrow\left(1\right)=t\left(t+24\right)+128=t^2+24t+128\)

\(=t^2+16t+8t+128\)

\(=t\left(t+16\right)+8\left(t+16\right)=\left(t+8\right)\left(t+16\right)\)(2)

Mà \(x^2+10x=t\)(ẩn phụ)

Nên \(\left(2\right)=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

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

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

Dấu = xảy ra khi x=4 hoặc x=2

\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)-7\)

\(=\left(x+1\right)\left(x+6\right)\left(x+3\right)\left(x+4\right)-7\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+12\right)-7\)

Đặt \(x^2+7x+9=t\)

\(=\left(t-3\right)\left(t+3\right)-7\)

\(=t^2-9-7=t^2-16=\left(t-4\right)\left(t+4\right)\)

\(=\left(x^2+7x+9-4\right)\left(x^2+7x+9+4\right)\)

\(=\left(x^2+7x+5\right)\left(x^2+7x+13\right)\)

Gợi ý:

a) Đặt  \(x^2+3x+1=a\)

b)  \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt     \(x^2+8x+11=a\)

c)  \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt    \(x^2+7x+11=a\)

d) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt   \(12x^2+11x-1=a\)

Câu hỏi của Nguyễn Tấn Phát - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo câu e nhé!

Hãy tích cho tui đi

vì ai tích cho tui thì người đó thông minh

ĐK:  \(-2\le x\le2\)

\(3\sqrt{2+x}-6\sqrt{2-x}+4\sqrt{4-x^2}=10-3x\)

<=>  \(3\left(\sqrt{2+x}-2\sqrt{2-x}\right)=10-3x-4\sqrt{4-x^2}\)

Đặt:  \(t=\sqrt{2+x}-2\sqrt{2-x}\)  =>   \(t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó pt trở thành:

\(3t=t^2\)

<=> \(t^2-3t=0\)

<=> \(t\left(t-3\right)=0\)

<=> \(\orbr{\begin{cases}t=0\\t=3\end{cases}}\)

đến đây bn tự giải nốt nhé

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

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

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

Đặt \(x^2-3x+1=t\)

\(=\left(t-1\right)\left(t+1\right)-3\)

\(=t^2-1-3=t^2-4\)

\(=\left(t-2\right)\left(t+2\right)\)

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

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