bổ sung:tìm điều kiện xác định của phương trình

ĐKXĐ : x khác cộng trừ 2

⇔ [( x + 2 )( x+12 )][( x + 3 )(x + 8)] = 4x²

⇔ ( x\(^2\) + 2x + 12x + 24 ) ( x\(^2\) + 3x + 8x + 24 ) = 4x²

Đặt x\(^2\) + 24 là a tacó :

pt⇔( a + 14x )( a + 11x ) = 4x\(^2\)

⇔ a\(^2\) + 11ax + 14ax + 154x\(^2\) - 4x\(^2\) = 0

⇔ a\(^2\) + 25ax + 150x\(^2\) = 0

⇔ a\(^2\) + 15ax + 10ax + 150x\(^2\) = 0

⇔ a( a + 15x ) + 10x ( a + 15x ) = 0

⇔ ( a + 10x ) ( a + 15x ) = 0

Thay a bằng x\(^2\) + 24

pt⇔ ( x\(^2\) + 24 + 10x ) ( x\(^2\) + 24 + 15x ) = 0

⇔ ( x\(^2\) + 4x + 6x + 24 ) ( x\(^2\) + 15x + 24 ) = 0

⇔ [ x( x + 4 ) + 6 (x + 4 )] ( np in dam) = 0

⇔ [ ( x + 6 ) ( x + 4 ) ] ( cnt ) = 0

⇔ \(\left[{}\begin{matrix}x+6=0\\x+4=0\\x^2+15x+24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-4\\x\approx-1,82\\x\approx-13,18\end{matrix}\right.\)

Viết đủ đề bài ra đi bạn, đề bài toàn thiếu vế phải

\(\Leftrightarrow\left(x+2\right)\left(x+12\right)\left(x+3\right)\left(x+8\right)=4x^2\)

\(\Leftrightarrow\left(x^2+14x+24\right)\left(x^2+11x+24\right)=4x^2\)

Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(\Leftrightarrow\left(x+14+\frac{24}{x}\right)\left(x+11+\frac{24}{x}\right)=4\)

Đặt \(x+11+\frac{24}{x}=a\)

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

\(\Rightarrow\left[{}\begin{matrix}x+11+\frac{24}{x}=1\\x+11+\frac{24}{x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+10x+24=0\\x^2+15x+24=0\end{matrix}\right.\) \(\Leftrightarrow...\)

(x+1)(x+2)(x+4)(x+8)=28x²

\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+9x+8\right)=28x^2\)(1)

Thấy x=0 không là nghiệm của (1). CHia 2 vế (1) cho x² ta đc:

\(\left(1\right)\Leftrightarrow\left(x+\frac{8}{x}+6\right)\left(x+\frac{8}{9}+9\right)=28\)

Đặt \(t=x+\frac{8}{x}\)ta có:

\(\left(1\right)\Rightarrow\left(t+6\right)\left(t+9\right)=28\)

\(\Leftrightarrow t^2+15t+26=0\Leftrightarrow\orbr{\begin{cases}t=-2\\t=-13\end{cases}}\)

\(\Delta=13^2-4\left(1.8\right)=137\)\(\Rightarrow x_{1,2}=\frac{-13\pm\sqrt{137}}{2}\)(thỏa mãn)

Vậy...

DK:....

\(\sqrt{1+x}+\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)

Dat \(\sqrt{1+x}+\sqrt{8-x}=p\)

\(\Leftrightarrow p^2=1+x+8-x+2\sqrt{\left(1+x\right)\left(8-x\right)}\)

\(\Leftrightarrow\frac{p^2-9}{2}=\sqrt{\left(1+x\right)\left(8-x\right)}\)

Ta co bien doi :

\(pt\Leftrightarrow p+\frac{p^2-9}{2}=3\)

\(\Leftrightarrow\frac{p^2+2p-9}{2}=3\)

\(\Leftrightarrow p^2+2p-15=0\)

\(\Leftrightarrow\left(p+5\right)\left(p-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}p=-5\\p=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}+\sqrt{8-x}=-5\left(loai\right)\\\sqrt{1+x}+\sqrt{8-x}=3\left(chon\right)\end{matrix}\right.\)

\(\Leftrightarrow1+x+8-x+2\sqrt{\left(1+x\right)\left(8-x\right)}=9\)

\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1+x=0\\8-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)( thoa )

Vay...

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

\(\Leftrightarrow4x^3=x^3-6x^2+12x-8\)

\(\Leftrightarrow4x^3=\left(x-2\right)^3\)

\(\Rightarrow\sqrt[3]{4}.x=x-2\)

\(\Rightarrow x=\dfrac{2}{1-\sqrt[3]{4}}\)

Lời giải:

PT \(\Leftrightarrow [(x-5)(x-8)][(x-4)(x-10)]=72x^2\)

\(\Leftrightarrow (x^2-13x+40)(x^2-14x+40)=72x^2\)

Đặt \(x^2-13x+40=a\) thì pt trở thành:

\(a(a-x)=72x^2\)

\(\Leftrightarrow a^2-ax-72x^2=0\)

\(\Leftrightarrow a^2-9ax+8ax-72x^2=0\)

\(\Leftrightarrow a(a-9x)+8x(a-9x)=0\)

\(\Leftrightarrow (a-9x)(a+8x)=0\)

Nếu $a-9x=0$

\(\Leftrightarrow x^2-13x+40-9x=0\)

\(\Leftrightarrow x^2-22x+40=0\)

\(\Leftrightarrow (x-2)(x-20)=0\Rightarrow \left[\begin{matrix} x=2\\ x=20\end{matrix}\right.\)

Nếu $a+8x=0$

\(\Leftrightarrow x^2-13x+40+8x=0\)

\(\Leftrightarrow x^2-5x+40=0\Leftrightarrow (x-\frac{5}{2})^2=-\frac{135}{4}\) (vô lý)

Vậy........

\(\dfrac{x+2}{89}+\dfrac{x+5}{86}>\dfrac{x+8}{83}+\dfrac{x+11}{80}\)

\(\Leftrightarrow\dfrac{x+2}{89}+1+\dfrac{x+5}{86}+1>\dfrac{x+8}{83}+1+\dfrac{x+11}{80}+1\)

\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)

\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}-\dfrac{x+91}{83}-\dfrac{x+91}{80}>0\)

\(\Leftrightarrow\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}-\dfrac{1}{83}-\dfrac{1}{80}\right)>0\)

Ta có: \(\dfrac{1}{89}+\dfrac{1}{86}+\dfrac{1}{83}+\dfrac{1}{80}< 0\)

\(\Leftrightarrow x+91< 0\)

\(\Leftrightarrow x< -91\)

Vậy...........