a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

Vậy...

mơn nha

Ta có :

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=1\\x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}}\)

Vậy \(x=-1\) hoặc \(x=\frac{1}{3}\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

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

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

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

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

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

Đặt \(\sqrt{x^2+2}=t>0\Rightarrow x^2=t^2-2\)

\(5\left(t^2-2\right)^2-2\left(t^2-2\right)-3t\left(t^2-2\right)-4=0\)

\(\Leftrightarrow5t^4-3t^3-22t^2+6t+20=0\)

Nhận thấy \(t=0\) không phải nghiệm, chia 2 vế cho \(t^2\)

\(\Rightarrow5\left(t^2+\frac{4}{t^2}\right)-3\left(t-\frac{2}{t}\right)-22=0\)

Đặt \(t-\frac{2}{t}=a\Rightarrow t^2+\frac{4}{t^2}=a^2+4\)

\(\Rightarrow5\left(a^2+4\right)-3a-22=0\Leftrightarrow5a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{2}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t-\frac{2}{t}=1\\t-\frac{2}{t}=-\frac{2}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t^2-t-2=0\\t^2+\frac{2}{5}t-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\\t=\frac{\sqrt{51}-1}{5}\\t=\frac{-\sqrt{51}-1}{5}\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=2\\\sqrt{x^2+2}=\frac{\sqrt{51}-1}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=\frac{2-2\sqrt{51}}{25}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=\pm\sqrt{2}\)

Bạn tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/giai-pt-sqrtx-2sqrt4-x2x2-5x-1.219493072549

đk: \(-x^4+3x-1\ge0\)

Có \(-\left(x^4+1\right)\le-2x^2\)

 \(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\) 

Áp dụng bunhia có: \(\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\le\sqrt{\left(1+1\right)\left(3x-2x^{^2}+2x^2-3x+2\right)}=2\)

\(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le2\)  (*)

Có: \(x^4-x^2-2x+4=\left(x^4+1\right)-x^2-2x+3\ge2x^2-x^2-2x+3=\left(x-1\right)^2+2\ge2\) (2*)

Từ (*) (2*) dấu = xảy ra khi x=1 (TM)

Vậy x=1

xét \(A=2x^2-2x-4=2\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\ge-\dfrac{9}{2}\)

\(\Rightarrow8^{2x^2-2x-4}\ge\dfrac{1}{\sqrt{8^9}}\)

Để phương trình: \(8^{2x^2-2x-4}+m^2-m=0\) có nghiệm

Cần \(m-m^2\ge\dfrac{1}{\sqrt{8^9}}\Leftrightarrow m^2-m+\dfrac{1}{\sqrt{.8^9}}\le0\)

\(\Rightarrow\dfrac{1-\sqrt{1-\dfrac{4}{\sqrt{8^9}}}}{2}\le m\le\dfrac{1+\sqrt{1-\dfrac{4}{\sqrt{8^9}}}}{2}\)

=>không có đáp án nào tuyệt đối chính xác.

chọn phương B gần đúng nhất nhưng vẫn chưa đúng:

do \(\left\{{}\begin{matrix}\dfrac{1+\sqrt{1-\dfrac{4}{\sqrt{8^9}}}}{2}< 1\\\dfrac{1-\sqrt{1-\dfrac{4}{\sqrt{8^9}}}}{2}>0\end{matrix}\right.\).

C

Lời giải:

Ta có \(8^{2x^2-2x-4}=m(1-m)\)

Ta biết rằng \(a^x>0\forall a>0\) nên \(8^{2x^2-2x-4}>0\forall x\)

\(\Rightarrow m(1-m)>0\). Giải BPT đó ta thu được \(\left[{}\begin{matrix}m< 0\\m>1\end{matrix}\right.\)

Đáp án C.