\(\dfrac{x}{2}\left(4x-3\right)+2\left(3-x\right)\left(x+4\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2}{2}-\dfrac{3x}{2}+2\left(3x+12-x^2-4x\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x}{2}+6x+24-2x^2-8x\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x+2\left(6x+24-2x^2-8x\right)}{2}\le0\)

\(\Leftrightarrow4x^2-3x+12x+48-4x^2-16x\le0\)

\(\Leftrightarrow-7x\le-48\)

\(\Leftrightarrow x\ge\dfrac{48}{7}\)

=>-7x+48≤0

<=>-7x≤-48

<=>(-7x)(-1)≥(-48)(-1)

<=>\(\dfrac{7x}{7}\)≥\(\dfrac{48}{7}\)

<=>x≥\(\dfrac{48}{7}\)

<=>x(x^2-1)=0

<=>x=0 hoặc x^2-1=0

<=>x=0 hoặc x^2=1

<=>x=0 hoặc x=1 hoặc x=-1

\(x^4-2x^3+4x^2-3x+2=0\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+\dfrac{1}{4}+\dfrac{5}{4}=1\\x^2-x+\dfrac{1}{4}+\dfrac{5}{4}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}=1\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=-\dfrac{1}{4}\\\left(x-\dfrac{1}{2}\right)^2=-\dfrac{9}{4}\end{matrix}\right.\)

\(\Rightarrow\) Vô lý ( vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\) )

\(\Rightarrow PT\) vô nghiệm .

\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)

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

\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)

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

\(\Leftrightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2=0\)

Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)

Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)

Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.

x^3-4x^2+5x-1-căn 2x-3=0

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

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

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

=>x-2=0

=>x=2

x^4-4x^3+5x^2-2x-20

=x^4-4x^3+4x^2+x^2-2x-20

=x^2(x^2-4x+4)+x^2-2x-20

=x^2(x-2)^2 + x^2-2x+1-21

=x^2(x-2)^2+(x-1)^2-21=0

<=>x^2(x-2)^2+(x-1)^2=21

từ đây bạn giải ra cx này phải đề là tìm nghiệm nguyên nhé :D

shitbo không biết làm thì thôi ...

\(x^4-4x^3+5x^2-2x-20=0\)

\(\Leftrightarrow\left(x^2-2x\right)^2+x^2-2x-20=0\)

Đặt \(x^2-2x=a\left(a\ge-1\right)\)

\(\Rightarrow pt:a^2+a-20=0\)

\(\Leftrightarrow\left(a-4\right)\left(a+5\right)=0\)

\(\Leftrightarrow a=4\left(Do\text{ }a\ge-1\right)\)

\(\Leftrightarrow x^2-2x=4\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}+1\\x=-\sqrt{5}+1\end{cases}}\)

Lần sau e sẽ rút kinh nghiệm thưa a :))

a)Dat \(x^2-4x+3=a;x^2-7x+6=b \Rightarrow a+b=2x^2-11x+9\)

....

a) ĐKXĐ: 1 ≥ x ≥ -1

Ta có: VT ≥ 0 = VP

Dấu "=" xảy ra khi và chỉ khi

\(\left\{{}\begin{matrix}\sqrt{1-x^2}=0\\\sqrt{1+x}=0\end{matrix}\right.\)

<=> x = -1 (TM)

b) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

Ta có: VT ≥ 0 = VP

Dấu "=" xảy ra khi và chỉ khi

\(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x^2+4x+4}=0\end{matrix}\right.\)

<=> x = -2 (TM)

c) \(\sqrt{1-x^2}+\sqrt{x+1}=0\)

ĐKXĐ: \(\left\{{}\begin{matrix}1-x^2\ge0\\x+1\ge0\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}1\ge x^2\\x\ge-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge-1\end{matrix}\right.\)

=> -1 \(\le\) x \(\le\) 1

\(\sqrt{1-x^2}+\sqrt{x+1}=0\)

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

\(\Leftrightarrow\)\(\sqrt{\left(1+x\right)}.\left(\sqrt{1-x}+1\right)=0\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{1+x}=0\\\sqrt{1-x}=-1\left(voli\right)\end{matrix}\right.\Rightarrow x+1=0\)

=> x = -1 ( thỏa mãn)

d) ĐKXĐ: \(x^2-4\ge0\Rightarrow x^2\ge4\)

\(\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

\(\sqrt{x^2-4}+\sqrt{\left(x+2^2\right)}=0\)

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

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+2=0\\\sqrt{x-2}=-\sqrt{x+2}\end{matrix}\right.\)

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

Vậy x= -2

Bạn tự phân tích đa thức thành nhân tử nhé! 

\(1.\)

\(2x^3+x+3=0\)

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

Vì  \(2x^2-2x+3=2\left(x^2-x+1\right)+1=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}>0\)  với mọi  \(x\in R\)

nên từ  \(\left(1\right)\)  \(\Rightarrow\)  \(x+1=0\)  \(\Leftrightarrow\)  \(x=-1\)

1)2x^3+x+3=0=>

Phân tích ra thôi bạn!