\(1,\Leftrightarrow x^2-8x+16-x^2+x+12=7\\ \Leftrightarrow-7x=-21\\ \Leftrightarrow x=3\\ 2,\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

IV: Để M nguyên thì \(3x^3-2x^2-6x+5\) ⋮3x-2

=>\(x^2\left(3x-2\right)-6x+4+1\) ⋮3x-2

=>1⋮3x-2

=>3x-2∈{1;-1}

=>3x∈{3;1}

=>x∈{1;1/3}

mà x nguyên

nên x=1

III:

1: \(\left(x-4\right)^2-\left(x+3\right)\left(x-4\right)=7\)

=>(x-4)(x-4-x-3)=7

=>-7(x-4)=7

=>x-4=-1

=>x=3

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

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

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

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=5\end{array}\right.\)

a: Trên tia đối của tia AC, lấy D sao cho AC=AD

=>A là trung điểm của CD

Xét ΔDBC có

BA là đường trung tuyến

\(BA=\frac{CD}{2}\)

Do đó: ΔDBC vuông tại B

Ta có: AH⊥BC

BD⊥BC

Do đó: AH//BD

Xét ΔBDC có AH//BD

nên \(\frac{AH}{BD}=\frac{CH}{CB}=\frac12\)

=>\(BD=2\cdot AH\)

Xet ΔBDC vuông tại B có BK là đường cao

nên \(\frac{1}{BK^2}=\frac{1}{BC^2}+\frac{1}{BD^2}=\frac{1}{BC^2}+\frac{1}{4\cdot AH^2}\)

c: BK//CJ

BK⊥AC

Do đó: AC⊥CJ

Xét ΔACJ vuông tại C có CI là đường cao

nên \(AI\cdot AJ=AC^2\)

=>\(AI\cdot AJ=AB^2\)

3.

\(4sinx+cosx+2cos\left(x+\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow4sinx+cosx+cosx-\sqrt{3}sinx=2\)

\(\Leftrightarrow\left(4-\sqrt{3}\right)sinx+2cosx=2\)

\(\Leftrightarrow\sqrt{23-4\sqrt{3}}\left(\dfrac{4-\sqrt{3}}{\sqrt{23-4\sqrt{3}}}sinx+\dfrac{2}{\sqrt{23-4\sqrt{3}}}cosx\right)=2\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}\right)=\dfrac{2}{\sqrt{23-4\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}=\pm arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

4.

\(sinx+2cos\left(x+\dfrac{\pi}{3}\right)+4sin\left(x+\dfrac{\pi}{6}\right)+cosx=4\)

\(\Leftrightarrow sinx+cosx-\sqrt{3}sinx+2\sqrt{3}sinx+2cosx+cosx=4\)

\(\Leftrightarrow\left(1+\sqrt{3}\right)sinx+4cosx=4\)

\(\Leftrightarrow\sqrt{20+2\sqrt{3}}\left(\dfrac{1+\sqrt{3}}{\sqrt{20+2\sqrt{3}}}sinx+\dfrac{4}{\sqrt{20+2\sqrt{3}}}cosx\right)=4\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}\right)=\dfrac{4}{\sqrt{20+2\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}=\pm arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

IV: Để M nguyên thì \(3x^3-2x^2-6x+5\) ⋮3x-2

=>\(x^2\left(3x-2\right)-6x+4+1\) ⋮3x-2

=>1⋮3x-2

=>3x-2∈{1;-1}

=>3x∈{3;1}

=>x∈{1;1/3}

mà x nguyên

nên x=1

III:

1: \(\left(x-4\right)^2-\left(x+3\right)\left(x-4\right)=7\)

=>(x-4)(x-4-x-3)=7

=>-7(x-4)=7

=>x-4=-1

=>x=3

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

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

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

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=5\end{array}\right.\)

1: \(3ab-6a^2b+3a^3b=3ab\cdot1-3ab\cdot2a+3ab\cdot a^2\)

\(=3ab\left(1-2a+a^2\right)=3ab\left(1-a\right)^2\)

2: \(x^2-7x-y^2-7y\)

\(=\left(x^2-y^2\right)-7\left(x+y\right)\)

=(x+y)(x-y)-7(x+y)

=(x+y)(x-y-7)

3: 5a(a-5)-2a+10

=5a(a-5)-2(a-5)

=(a-5)(5a-2)

4: \(5x\left(x-3\right)-x^2+9\)

=5x(x-3)-(x-3)(x+3)

=(x-3)(5x-x-3)

=(4x-3)(x-3)

5: \(9a^2-b^2+4b-4\)

\(=\left(3a\right)^2-\left(b^2-4b+4\right)\)

\(=\left(3a\right)^2-\left(b-2\right)^2\)

=(3a-b+2)(3a+b-2)

6: \(2x^2-x-6\)

\(=2x^2-4x+3x-6\)

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

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

7: \(6x^3-15x^2=3x^2\cdot2x-3x^2\cdot5=3x^2\cdot\left(2x-5\right)\)

8: \(9x^2-25=\left(3x\right)^2-5^2=\left(3x-5\right)\left(3x+5\right)\)

9: \(4x^3-4x^2y-x+xy^2\)

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

\(=x\left\lbrack\left(2x-y\right)^2-1\right\rbrack\)

=x(2x-y-1)(2x-y+1)

IV: Để M nguyên thì \(3x^3-2x^2-6x+5\) ⋮3x-2

=>\(x^2\left(3x-2\right)-6x+4+1\) ⋮3x-2

=>1⋮3x-2

=>3x-2∈{1;-1}

=>3x∈{3;1}

=>x∈{1;1/3}

mà x nguyên

nên x=1

III:

1: \(\left(x-4\right)^2-\left(x+3\right)\left(x-4\right)=7\)

=>(x-4)(x-4-x-3)=7

=>-7(x-4)=7

=>x-4=-1

=>x=3

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

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

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

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=5\end{array}\right.\)

1: Gọi ba số tự nhiên đề bài cho là a,b,c

(ĐIều kiện: a<>b; b<>c; a<>c)

Tổng của chúng bằng tích của chúng nên ta có: abc=a+b+c

=>a=1; b=2; c=3

Bài 2:

\(x\left(x+1\right)=2+4+6+\cdots+2500\)

=>\(x\left(x+1\right)=2\left(1+2+\cdots+1250\right)\)

=>\(x\left(x+1\right)=2\cdot\frac{1250\cdot1251}{2}=1250\cdot1251\)

=>\(x^2+x-1250\cdot1251=0\)

=>(x+1251)(x-1250)=0

=>x=-1251 hoặc x=1250