a) Đúng

b)Đúng

c)Sai vì nghiệm không thỏa mãn ĐKXĐ

d)Sai vì có 1 nghiệm không thỏa mãn ĐKXĐ

\(x^2+6x+9=\left(x+3\right)^2\)

--

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

--

\(x^3+12x^2+48x+64=\left(x+4\right)^3\)

1) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)

\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)

\(=\dfrac{2x^2+50}{x^2+25}\)

\(=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\)

2) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3+3^3-54-x^3\)

\(=27-54=-27\)

3) \(\left(2x+y\right)^2-\left(y+3x\right)^2\)

\(=4x^2+4xy+y^2-y^2-6xy-9x^2\)

\(=-5x^2-2xy\)

4) \(\left(2x+1\right)^3-\left(2x-1\right)^3-24x^2\)

\(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2\)

\(=2\)

1)

ĐK: \(x,y\neq 0\); \(x+y\neq 0\)

\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)

\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)

2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)

\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)

\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)

3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)

\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)

4) ĐK: \(x\neq \frac{\pm 1}{3}\)

\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)

\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)

\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)

5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)

\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)

\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)

\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)

\(=\frac{3}{(x+1)^2}\)

Gợi ý:

a) Đặt  \(x^2+3x+1=a\)

b)  \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt     \(x^2+8x+11=a\)

c)  \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt    \(x^2+7x+11=a\)

d) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt   \(12x^2+11x-1=a\)

Câu hỏi của Nguyễn Tấn Phát - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo câu e nhé!

a.

$4(x+5)(x+6)(x+10)(x+12)=3x^2$

$4[(x+5)(x+12)][(x+6)(x+10)]=3x^2$

$4(x^2+17x+60)(x^2+16x+60)=3x^2$

Đặt $x^2+16x+60=a$ thì pt trở thành:

$4(a+x)a=3x^2$

$4a^2+4ax-3x^2=0$

$4a^2-2ax+6ax-3x^2=0$

$2a(2a-x)+3x(2a-x)=0$

$(2a-x)(2a+3x)=0$

Nếu $2a-x=0\Leftrightarrow 2(x^2+16x+60)-x=0$

$\Leftrightarrow 2x^2+31x+120=0\Rightarrow x=\frac{-15}{2}$ hoặc $x=-8$

Nếu $2a+3x=0\Leftrightarrow 2(x^2+16x+60)+3x=0$

$\Leftrightarrow 2x^2+35x+120=0\Rightarrow x=\frac{-35\pm \sqrt{265}}{4}$

b.

$(x+1)(x+2)(x+3)(x+6)=120x^2$

$[(x+1)(x+6)][(x+2)(x+3)]=120x^2$

$(x^2+7x+6)(x^2+5x+6)=120x^2$

Đặt $x^2+6=a$ thì pt trở thành:

$(a+7x)(a+5x)=120x^2$

$\Leftrightarrow a^2+12ax-85x^2=0$

$\Leftrightarrow a^2-5ax+17ax-85x^2=0$

$\Leftrightarrow a(a-5x)+17x(a-5x)=0$

$\Leftrightarrow (a-5x)(a+17x)=0$

Nếu $a-5x=0\Leftrightarrow x^2+6-5x=0$

$\Leftrightarrow (x-2)(x-3)=0\Rightarrow x=2$ hoặc $x=3$

Nếu $a+17x=0\Leftrightarrow x^2+17x+6=0$

$\Rightarrow x=\frac{-17\pm \sqrt{265}}{2}$

Vậy.........

a, x³ +x² -12x=0

\(\Leftrightarrow\)x³ +4x²-3x²-12x=0

\(\Leftrightarrow\) x²(x+4)-3x(x+4)=0

\(\Leftrightarrow\) (x²-3x)(x+4)=0

\(\Leftrightarrow\)x(x-3)(x+4)=0

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

Vậy S\(=\)\(\left\{0;3;-4\right\}\)

b.x³-4x²-x+4=0

\(\Leftrightarrow\)x²(x-4)-(x-4)=0

\(\Leftrightarrow\) (x² -1)(x-4)=0

\(\Leftrightarrow\)(x-1)(x+1)(x-4)=0

\(\left[\begin{matrix}x+1=0\\x-1=0\\x-4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=1\\x=-1\\x=4\end{matrix}\right.\)

Vậy S=\(\left\{1;-1;4\right\}\)

nếu từ trên mà bạn suy ra xuống dưới thì đúng