Bài 4: Phương trình tích

Ta có : \(x^2-4-5\left(x-2\right)^2=0\)

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

\(\Leftrightarrow\left(x-2\right)\left[x+2-5\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-5x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(12-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\12-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\-4x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;3\right\}\)

Ta có: \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)

\(a,x^4-16x^2+32x-16=0\)

\(\Leftrightarrow\left(x^4-16\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^4+4\right)\left(x-2\right)\left(x+2\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x-4x+8\right)=0\)\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)

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

\(\Leftrightarrow\left(x-2\right)^2\left[\left(x+2\right)^2-8\right]=0\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{8}\\x+2=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}-2\\x=-\sqrt{8}-2\end{matrix}\right.\)

câu nào dễ xơi trước

g) \(x^3+3x^2-2x-6=0\Leftrightarrow x^2\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x+3\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}\\x=-3\end{matrix}\right.\)

kl: ...........

Đề bài sai

Đặt bt trong ngoặc đầu tiên = t

pt trở thành

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

với t=3, ta có:

\(x^2+2x-1=3\Leftrightarrow x^2+2x-4=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)

t= -1 tương tự

1) Gọi x(km/h) x>0 là vận tốc bình thường

Xuôi dòng: x+3(km/h)

Ngược dòng : x-3(km/h)

Quãng đường biểu thị khi xuôi dòng:

\(\left(x+3\right)1,5\left(km\right)\)

Quãng đường biểu thị khi ngược dòng:

\(\left(x-3\right)2\left(km\right)\)

Theo đề, ta có pt:

\(\left(x+3\right)1,5=\left(x-3\right)2\)

\(1,5x+4,5=2x-6\)

\(2x-1,5x=6+4,5\)

\(0,5x=10,5\)

\(x=21\left(TMĐK\right)\)

Quãng đường là: (21+3).1,5=36(km)

Thiếu mấy dấu ''<=>'' nha

theo bài ra ta có:

\(\dfrac{\left(2x+7\right)^2}{9}=\left(x+2\right)^2\)

\(=>\left(\dfrac{2x+7}{3}\right)^2=\left(x+2\right)^2\)

\(=>\dfrac{2x+7}{3}=x+2\)

giải ra ta tìm được x=1

CHÚC BẠN HỌC TỐT............

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+23x+35x+115=0\)

\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\Leftrightarrow x=-5\\7x+23=0\Leftrightarrow x=-\dfrac{23}{7}\end{matrix}\right.\)

Vậy \(S=\left\{-5;-\dfrac{23}{7}\right\}\)

a)x⁴-4x³+12x-9 = x³(x-1) -3x²(x-1) -3x(x-1) +9(x-1)

=(x-1)(x³-3x²-3x+9)

=(x-1)[x²(x-3)-3(x-3)]

=(x-1)(x-3)(x²-3)

b)(x+2)²=9(x²-4x+4) <--> x²+4x+4=9x²-36x+36

<-->8x² -40x+32=0

<-->8(x²-5x+4)=0

<-->x²-5x+4=0

<--->(x-4)(x-1)=0

* Nếu x-4=0 <--> x=4

* Nếu x-1=0<--> x=1

Vậy S={4;1}