Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$

$\Leftrightarrow 2+2\sqrt{(x-2)(4-x)}=2$
$\Leftrightarrow (x-2)(4-x)=0$

$\Leftrightarrow x-2=0$ hoặc $4-x=0$

$\Leftrightarrow x=2$ hoặc $x=4$ (tm)

Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$

$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$

Với $4x^3-3x^2+6x-4=0(*)$

Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$

Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:

$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$

Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)

Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)

Nãy mình tìm được một cách giải tương tự cho câu 2.

PT \(\Leftrightarrow\left(x-1\right)\left(4x^3-3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x^3-3x^2+6x-4=0\left(1\right)\end{matrix}\right.\)

Vậy pt có 1 nghiệm bằng 1.

\(\left(1\right)\Rightarrow8x^3-6x^2+12x-8=0\)

\(\Leftrightarrow7x^3+x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=-7x^3\)

\(\Leftrightarrow x-2=-\sqrt[3]{7}x\)

\(\Leftrightarrow x=\dfrac{2}{1+\sqrt[3]{7}}\)

Vậy pt có nghiệm \(S=\left\{1;\dfrac{2}{1+\sqrt[3]{7}}\right\}\)

Lưu ý: Nghiệm của người kia hoàn toàn tương đồng với nghiệm của mình (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

mấy chế nhanh giúp mik vs

Đặt \(\left(x^2-x+1\right)^2=a;x^2=b\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a^2-10ab+9b^2=0\\ \Leftrightarrow a^2-9ab-ab+9b^2=0\\ \Leftrightarrow\left(a-b\right)\left(a-9b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\a=9b\end{matrix}\right.\\ \forall a=b\Leftrightarrow\left(x^2-x+1\right)^2-x^2=0\\ \Leftrightarrow\left(x^2-2x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow x=1\\ \forall a=9b\Leftrightarrow\left(x^2-x+1\right)^2-9x^2=0\\ \Leftrightarrow\left(x^2-4x+1\right)\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

x4+10x3+26x2+10x+1=0x4+10x3+26x2+10x+1=0

⇔x4+6x3+x2+4x3+24x2+4x+x2+6x+1=0⇔x4+6x3+x2+4x3+24x2+4x+x2+6x+1=0

⇔x2(x2+6x+1)+4x(x2+6x+1)+(x2+6x+1)=0⇔x2(x2+6x+1)+4x(x2+6x+1)+(x2+6x+1)=0

⇔(x2+4x+1)(x2+6x+1)=0⇔(x2+4x+1)(x2+6x+1)=0

⇔(x2+4x+4−3)(x3+6x+9−8)=0⇔(x2+4x+4−3)(x3+6x+9−8)=0

⇔[(x+2)2−3][(x+3)2−8]=0⇔[(x+2)2−3][(x+3)2−8]=0

⇒[(x+2)2−3=0(x+3)2−8=0⇒[(x+2)2−3=0(x+3)2−8=0⇒[(x+2)2=3(x+3)2=8⇒[(x+2)2=3(x+3)2=8⇒⎡⎣⎢⎢⎢x=−4±12−−√2x=−6±32−−√2

Thử phân tích VT thành: \(\left(x^2+6x+1\right)\left(x^2+4x+1\right)=0\) xem sao?

\(x^4+10x^3+26x^2+10x+1=0\)

\(\Leftrightarrow\left(x^4+6x^3+x^2\right)+\left(4x^3+24x^2+4x\right)+\left(x^2+6x+1\right)=0\)

\(\Leftrightarrow x^2\left(x^2+6x+1\right)+4x\left(x^2+6x+1\right)+\left(x^2+6x+1\right)=0\)

\(\Leftrightarrow\left(x^2+6x+1\right)\left(x^2+4x+1\right)=0\)

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

\(\Leftrightarrow\left[\left(x+3\right)^2-8\right]\left[\left(x+2\right)^2-3\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2-8=0\\\left(x+2\right)^2-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=8\\\left(x+2\right)^2=3\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\pm\sqrt{8}\\x+2=\pm\sqrt{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{8}-3\\x=\pm\sqrt{3}-2\end{cases}}}\)

a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)

\(\Leftrightarrow\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-12=0\)

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

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+5\right)=0\)

\(\Leftrightarrow x^2+x+4=0\) hay \(x^2+x-2=0\)

\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{15}{4}=0\) hay \(x^2-x+2x-2=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) (pt vô nghiệm) hay\(x\left(x-1\right)+2\left(x-1\right)=0\)

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

\(\Leftrightarrow x=1\) hay \(x=-2\)

-Vậy \(S=\left\{1;-2\right\}\)

b) \(x^3+5x^2-10x-8=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+4x-8=0\)

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

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

\(\Leftrightarrow x=2\) hay \(x^2+2.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{33}{4}=0\)

\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{33}{4}=0\)

\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{7}{2}-\dfrac{\sqrt{33}}{2}\right)=0\)

\(\Leftrightarrow x=2\) hay \(x=\dfrac{-7-\sqrt{33}}{2}\) hay \(x=\dfrac{-7+\sqrt{33}}{2}\)

-Vậy \(S=\left\{2;\dfrac{-7-\sqrt{33}}{2};\dfrac{-7+\sqrt{33}}{2}\right\}\)

1. phương trình tương đương với \(\left(x^2-7x+2\right)\left(x^2+2x+2\right)=0\to x=\frac{7}{2}\pm\frac{\sqrt{41}}{2}\)

2. phương trình tương đương với \(\left(x^2+\left(\sqrt{2}-1\right)x+1\right)\left(x^2+\left(\sqrt{2}+1\right)x-1\right)=0\to x=\frac{-1\pm\sqrt{2}\pm\sqrt{7-2\sqrt{2}}}{2}\) với dấu +,- lấy tuỳ ý

Quan trọng là cách làm kìa. Chứ bấm máy là nghề của anh

          \(x^4-10x^3+26x^2-10x+1=0\)

\(\Leftrightarrow\)\(\left(x^4-4x^3+x^2\right)-\left(6x^3-24x+6x\right)+\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow\)\(x^2\left(x^2-4x+1\right)-6x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x^2-6x+1\right)\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2-6x+1=0\\x^2-4x+1=0\end{cases}}\)

Nếu   \(x^2-6x+1=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3-\sqrt{8}\\x=\sqrt{8}+3\end{cases}}\)

Nếu  \(x^2-4x+1=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2-\sqrt{3}\\x=\sqrt{3}+2\end{cases}}\)

Vậy....