Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)

= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

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

có  tính 6a ko

kick    tui  đi

mk học lp 6B nhưng là trường THCS Trung Hòa

Tui tự vẽ đó

la cau hoi tuong tu do ban

câu hỏi tương tự nha 

bằng 1,06 nha không dư tick giumg nha

bài này không cần giảng bạn ạ

Trang nè kết quả là 1,06

**** mình nha Trang xinh đẹp

284:

a: \(\left(x^2-6x+9\right)^2-15\left(x^2-6x+10\right)=1\)

=>\(\left(x^2-6x+9\right)^2-15\left(x^2-6x+9+1\right)-1=0\)

=>\(\left(x^2-6x+9\right)^2-15\left(x^2-6x+9\right)-16=0\)

=>\(\left(x^2-6x+9-16\right)\left(x^2-6x+9+1\right)=0\)

=>\(\left\lbrack\left(x-3\right)^2-16\right\rbrack\left\lbrack\left(x-3\right)^2+1\right\rbrack=0\)

=>\(\left(x-3\right)^2-16=0\)

=>(x-3-4)(x-3+4)=0

=>(x-7)(x+1)=0

=>\(\left[\begin{array}{l}x-7=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-1\end{array}\right.\)

b: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

=>\(\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2=0\)

=>\(\left(x^2+x+1\right)\left(x^2+2x+1\right)=0\)

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34>0\forall x\)

nên \(x^2+2x+1=0\)

=>\(\left(x+1\right)^2=0\)

=>x+1=0

=>x=-1

c: \(\left(x^2-9\right)^2=12x+1\)

=>\(x^4-18x^2+81-12x-1=0\)

=>\(x^4-18x^2-12x+80=0\)

=>\(x^4-6x^3+8x^2+6x^3-36x^2+48x+10x^2-60x+80=0\)

=>\(\left(x^2-6x+8\right)\cdot\left(x^2+6x+10\right)=0\)

mà \(x^2+6x+10=\left(x+3\right)^2+1\ge1>0\forall x\)

nên \(x^2-6x+8=0\)

=>(x-2)(x-4)=0

=>x=2 hoặc x=4

Bài 282:

a: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

=>\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)=0\)

=>(x-1)(x-4)(x-2)(x-3)=0

=>x∈{1;4;2;3}

b: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)

=>\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)

=>\(\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)

=>(x+6)(x-1)(x+4)(x+1)=0

=>x∈{-6;1;-4;-1}

c: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

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

=>\(\left(x^2+x\right)^2+3\left(x^2+x\right)-10=0\)

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

mà \(x^2+x+5=\left(x+\frac12\right)^2+\frac{19}{4}>0\forall x\)

nên \(x^2+x-2=0\)

=>(x+2)(x-1)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=1\end{array}\right.\)

d: \(\left(x^2+x-2\right)\left(x^2+x-3\right)=12\)

=>\(\left(x^2+x\right)^2-5\left(x^2+x\right)+6-12=0\)

=>\(\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)

=>\(\left(x^2+x-6\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34>0\forall x\)

nên \(x^2+x-6=0\)

=>(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

e: x(x+1)\(\left(x^2+x+1\right)=42\)

=>\(\left(x^2+x\right)\left(x^2+x+1\right)=42\)

=>\(\left(x^2+x\right)^2+\left(x^2+x\right)-42=0\)

=>\(\left(x^2+x+7\right)\left(x^2+x-6\right)=0\)

mà \(x^2+x+7=\left(x+\frac12\right)^2+\frac{27}{4}\ge\frac{27}{4}>0\forall x\)

nên \(x^2+x-6=0\)

=>(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

Với \(x\ge1\)

\(\sqrt{x+4}-\sqrt{x-1}=1\)

<=>\(\sqrt{x+4}=\sqrt{x-1}+1\)

<=>\(x+4=x-1+1+2\sqrt{x-1}\)

<=>\(2\sqrt{x-1}=4\)

<=>\(\sqrt{x-1}=2\)

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

<=>x=5(TM)

bước tương đương thứ 2 bình phương cả 2 vế

51/ a. did                       b. got                    c. made                          d. bought

52/ a. in                         b. from                 c. by                                d. of

53/ a. said                     b. asked                c. spoke                          d. told

54/ a. special                 b. specially           c. specialty                      d. specialty

55/ a. be                        b. are                    c. was                              d. were

56/ a. fashion                b. fashionable        c. fashionably                 d. fashioners

57/ a see                        b. saw                   c. seen                            d. seeing

58/ a. of                        b. to                      c. an                                d. a

59/ a. out                      b. of                      c. out of                          d. from

60/ a. with                     b. by                     c. of                               d. in