\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)

\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)

Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)

\(\Rightarrow4ab=\left(a+b\right)^2\)

\(\Leftrightarrow4ab=a^2+a^2+2ab\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow x^4-16x-12=0\)

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

\(\Leftrightarrow x^2-2x-2=0\)

\(\Rightarrow x=1\pm\sqrt{3}\)

\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)

\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)

\(x^4-8x^3+21x^2-24x+9=0\)

\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))

\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)

\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)

\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)

\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)

\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)

\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)

Làm tiếp bài của Song Ngư (๖ۣۜO๖ۣۜX๖ۣۜA)

\(D=\frac{\frac{-188}{11}x^{17}y^3}{6z^4a}\)

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

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

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

a) \(5x\left(3x-7\right)-15x\left(x-1\right)=3\)

\(\Rightarrow15x^2-35x-15x^2+15x=3\)

\(\Rightarrow-20x=3\)

\(\Rightarrow x=-\dfrac{3}{20}\)

b) \(\left(4x+2\right)\left(6x-3\right)-\left(8x+5\right)\left(3x-4\right)=2\)

\(\Rightarrow24x^2+12x-12x-6-24x^2-15x+24x+20=2\)

\(\Rightarrow9x+14=2\)

\(\Rightarrow9x=-12\)

\(\Rightarrow x=-\dfrac{4}{3}\)

c) \(7x^2-21x=0\)

\(\Rightarrow7x\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

d) \(9x^2-6x+1=0\)

\(\Rightarrow\left(3x\right)^2-2.3x+1=0\)

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

\(\Rightarrow3x-1=0\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\dfrac{1}{3}\)

e) \(16x^2-49=0\)

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

\(\Rightarrow\left(4x-7\right)\left(4x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}4x-7=0\\4x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=7\\4x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)

f) \(5x^3-20x=0\)

\(\Rightarrow5x\left(x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=0\\x^2-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x^2=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)

Casio:

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

b/ \(\Leftrightarrow2\left(2x^2+3x+3\right)^2+6\left(x+\frac{2}{3}\right)^2+\frac{28}{3}=0\)

Vế trái luôn dương nên pt vô nghiệm

c/ Câu này đề sai, pt này ko thể tách ra được nên chắc chắn là ko giải được

d/ Câu này chắc đề cũng ko đúng: đặt \(2x-4=a\Rightarrow2x=a+4\)

\(\Rightarrow\left(a+5\right)\left(a+1\right)\left(a+2\right)\left(a+10\right)=100\)

\(\Leftrightarrow a\left(a^3+18a^2+97a+180\right)=0\)

Dù pt có nghiệm \(a=0\) nhưng pt bậc 3 đằng sau lại ko thể giải

e/ Câu này giống câu trên

\(\Leftrightarrow x\left(16x^3+16x^2-93x+12\right)=0\)

Pt bậc 3 phía sau ko giải được