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Lời giải:
Ta có:
\((x+3)(x+12)(x-4)(x-16)+20x^2=0\)
\(\Leftrightarrow [(x+3)(x-16)][(x+12)(x-4)]+20x^2=0\)
\(\Leftrightarrow (x^2-13x-48)(x^2+8x-48)+20x^2=0\)
Đặt \(x^2-12x-48=a\). PT trở thành:
\((a-x)(a+20x)+20x^2=0\)
\(\Leftrightarrow a^2+19ax-20x^2+20x^2=0\Leftrightarrow a^2+19ax=0\)
\(\Leftrightarrow a(a+19x)=0\)
\(\Leftrightarrow (x^2-12x-48)(x^2+7x-48)=0\)
\(\Leftrightarrow \left[\begin{matrix} x^2-12x-48=0\\ x^2+7x-48=0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=6\pm 2\sqrt{21}\\ x=\frac{-7\pm \sqrt{241}}{2}\end{matrix}\right.\)
Vậy......
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
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 \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)
\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)
\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)
\(x^4-8x^3+21x^2-24x+9=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))
\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)
a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
Vậy x = 8 hoặc x = -7
a: Ta có: \(x^4-x^2-56=0\)
\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)
\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)
\(\Leftrightarrow x^2-8=0\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
cả 2 pt đều giải theo kiểu cái đầu nhóm với cái cuối, 2 cái ở giữa nhóm với nhau. sau đó giải theo cách đặt ẩn phụ
1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24=0\)
\(\Leftrightarrow\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
Đặt \(x^2+7x=a\), nên ta có :
\(\left(a+10\right)\left(a+12\right)-24=0\)
\(\Leftrightarrow\left(x+11-1\right)\left(x+11+1\right)-24=0\)
\(\Leftrightarrow\left[\left(x+11\right)^2-1\right]-24=0\)
\(\Leftrightarrow\left(x+11\right)^2-25=0\)
\(\Leftrightarrow\left(x+11-5\right)\left(x+11+5\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+16\right)=0\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-16\end{cases}}\)
Lần sau đừng tự tiện xếp vào phần bất pt bạn nhé :(
Ta có : \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)=3x^2\)
\(\Leftrightarrow4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)=3x^2\)
\(\Leftrightarrow4\left(x^2+17x+60\right)\left(x^2+16x+60\right)=3x^2\)(1)
Đặt \(x^2+16x+60=a\)
Pt (1) \(\Leftrightarrow4\left(a+x\right)a=3x^2\)
\(\Leftrightarrow4\left(a^2+ax\right)=3x^2\)
\(\Leftrightarrow4a^2+4ax=3x^2\)
\(\Leftrightarrow4a^2+4ax+x^2=4x^2\)
\(\Leftrightarrow\left(2a+x\right)^2=4x^2\)
\(\Leftrightarrow\orbr{\begin{cases}2a+x=2x\\2a+x=-2x\end{cases}}\)
*Nếu \(2a+x=2x\)
\(\Leftrightarrow2a=x\)
\(\Leftrightarrow x^2+16x+60=x\)
\(\Leftrightarrow x^2+15x+60=0\)
\(\Leftrightarrow x^2+2.\frac{15}{2}.x+\frac{225}{4}+\frac{15}{4}=0\)
\(\Leftrightarrow\left(x+\frac{15}{2}\right)^2+\frac{15}{4}=0\)
Pt vô nghiệm
*Nếu \(2a+x=-2x\)
\(\Leftrightarrow2a+3x=0\)
\(\Leftrightarrow2\left(x^2-16x+60\right)+3x=0\)
\(\Leftrightarrow2x^2-32x+120+3x=0\)
\(\Leftrightarrow2x^2-29x+120=0\)
\(\Leftrightarrow x^2-\frac{29}{2}x+60=0\)
\(\Leftrightarrow x^2-2.\frac{29}{4}.x+\frac{841}{16}+\frac{119}{16}=0\)
\(\Leftrightarrow\left(x-\frac{29}{4}\right)^2+\frac{119}{16}=0\)
Pt vô nghiệm
Vậy pt vô nghiệm