a: Ta có: ΔOCK vuông tại C

=>\(CK^2+CO^2=OK^2\)

=>\(CK=\sqrt{10^2-8^2}=6\left(cm\right)\)

Xét ΔOCK vuông tại C có CA là đường cao

nên \(OA\cdot OK=OC^2;CA\cdot OK=CO\cdot CK\)

=>\(OA=\dfrac{OC^2}{OK}=\dfrac{8^2}{10}=6,4\left(cm\right);CA=\dfrac{6\cdot8}{10}=4,8\left(cm\right)\)

Xét ΔCOK vuông tại C có \(sinCOK=\dfrac{CK}{OK}=\dfrac{6}{10}=\dfrac{3}{5}\)

nên \(\widehat{COK}=\widehat{xOy}\simeq36^052'\)

b: Xét ΔCAO vuông tại A có AH là đường cao

nên \(CH\cdot CO=CA^2\left(1\right)\)

Xét ΔCOK vuông tại O có CA là đường cao

nên \(AO\cdot AK=AC^2\left(2\right)\)

Từ (1),(2) suy ra \(CH\cdot CO=AO\cdot AK\)

Tớ sẽ làm mẫu cho cậu 1 số bài nhé:

a) \(A=\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{25-24}{24.25}\)

   \(A=\dfrac{6}{5.6}-\dfrac{5}{5.6}+\dfrac{7}{6.7}-\dfrac{6}{6.7}+...+\dfrac{25}{24.25}-\dfrac{24}{24.25}\)

   \(A=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

   \(A=\dfrac{1}{5}-\dfrac{1}{25}\)

   \(A=\dfrac{4}{25}\)

Bài 4; 2 và câu d bài 1 cậu sẽ cần phải đưa tử về hiệu giữa 2 thừa số ở mẫu.

\(\dfrac{4}{5}\) K = \(\dfrac{7-3}{3.7}+\dfrac{11-7}{7.11}+\dfrac{15-11}{11.15}=...+\dfrac{85-81}{81.85}+\dfrac{89-85}{85.89}\)

\(\dfrac{4}{5}K=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{81}-\dfrac{1}{85}+\dfrac{1}{85}-\dfrac{1}{89}\)

\(\dfrac{4}{5}K=\dfrac{1}{3}-\dfrac{1}{89}\)

\(\dfrac{4}{5}K=\dfrac{43}{147}\)

\(K=\dfrac{43}{147}\div\dfrac{4}{5}\)

\(K=\dfrac{215}{588}\)

Với bài 3 thì cậu chỉ cần đảo vị trí từ dưới lên trên là được nhé.

Bài 5: (Viết lại tổng E). Khoảng cách giữa 2 thừa số ở mẫu là 6, cậu hãy nhân tử với 6, tính sau đó : 6 nhé.

→ E = \(\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

Bài 6. Quan sát:

\(3\left(\dfrac{1}{5}-\dfrac{3}{x-2}\right)=\dfrac{24}{35}\) và tương tự như câu b, luôn là cái đầu tiên - cái cuối cùng.

Bài 7. Cậu trừ 1 ở cả 2 vế rồi nhân \(\dfrac{1}{2}\).

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\)

= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\)

Cậu cứ làm như những bài trên nhé.

Bài 1: 

\(a,A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=\dfrac{1}{5}-\dfrac{1}{25}=>\dfrac{5}{25}-\dfrac{1}{25}\)

\(=\dfrac{4}{25}\)

\(b,B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

       \(=1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

       \(=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

       \(=1.\left(1-\dfrac{1}{101}\right)\)

       \(=\dfrac{100}{101}\)

\(c,K=\dfrac{4}{11.16}+\dfrac{4}{16.21}+\dfrac{4}{21.26}+...+\dfrac{4}{61.66}\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11.16}+\dfrac{1}{16.21}+\dfrac{1}{21.26}+...+\dfrac{1}{61.66}\right)\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\right)\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11}-\dfrac{1}{66}\right)\)

       \(=\dfrac{4}{5}.\dfrac{5}{66}=>4.\dfrac{1}{66}\)

       \(=\dfrac{4}{66}=\dfrac{2}{33}\)

\(d,N=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)

        \(=2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

        \(=2.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

        \(=2.\left(1-\dfrac{1}{101}\right)\)

        \(=2.\dfrac{100}{101}\)

        \(=\dfrac{200}{101}\)

Bài 2:

\(K=\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{81.85}+\dfrac{5}{85.89}\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{81.85}+\dfrac{1}{85.89}\right)\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{89}\right)\)

    \(=\dfrac{5}{4}.\dfrac{86}{267}\)

    \(=\dfrac{215}{534}\)

Bài 3:

\(A=\dfrac{1}{25.24}+\dfrac{1}{24.23}+...+\dfrac{1}{7.6}+\dfrac{1}{6.5}\)

    \(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)

    \(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

    \(=\dfrac{1}{5}-\dfrac{1}{25}\)

    \(=\dfrac{4}{25}\)

Bài 4 :

\(A=\dfrac{5}{3.6}+\dfrac{5}{6.9}+\dfrac{5}{9.12}+...+\dfrac{5}{99.102}\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{99.102}\right)\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{102}\right)\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3}-\dfrac{1}{102}\right)\)

    \(=\dfrac{5}{3}.\dfrac{11}{34}\)

    \(=\dfrac{55}{102}\)

Bài 5 :

Sửa đề :\(a,E=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

        \(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

        \(=\dfrac{1}{6}.\left(\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\right)\)

        \(=\dfrac{1}{6}.\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

        \(=\dfrac{1}{6}.\left(1-\dfrac{1}{37}\right)\)

        \(=\dfrac{1}{6}.\dfrac{36}{37}\)

        \(=\dfrac{6}{37}\)

\(b,C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)

       \(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

       \(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

       \(=\dfrac{1}{3}-\dfrac{1}{13}\)

       \(=\dfrac{10}{39}\)

Bài 6 :

\(a,\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\) 

    \(\dfrac{3}{2}\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)

    \(\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)

          \(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)

          \(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)

                  \(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)

                  \(\dfrac{1}{x+2}=-\dfrac{9}{35}\)

                  \(-9\left(x+2\right)=1.35\)

                  \(-9\left(x+2\right)=35\)

                         \(x+2=35:-9\)

                        \(x+2=\dfrac{-35}{9}\)

                        \(x\)        \(=\dfrac{-35}{9}-2\)

                        \(x\)        \(=\dfrac{-53}{9}\)

Vậy \(x=\dfrac{-53}{9}\)

\(b,\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{10.13}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{1}{4}-\dfrac{1}{x+3}\right)\)                                        \(=\dfrac{1}{9}\)

   \(\dfrac{1}{6}-\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{9}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{6}-\dfrac{1}{9}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{18}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{2}{36}\)

      ⇒   \(3.\left(x+3\right)=36\)

                 \(x+3=36:3\)

                 \(x+3=12\) 

                 \(x\)       \(=12-3\)

                 \(x\)       \(=9\)

Vậy \(x=9\)

Bài 7:

\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)

\(=>\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{3980}{1991}\)

\(=>\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{3980}{1991}\)

\(=>2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(=>2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(=>2.\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

             \(1-\dfrac{1}{x+1}=\dfrac{3980}{1991}:2\)

             \(1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

                    \(\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)

                    \(\dfrac{1}{x+1}=\dfrac{1}{1991}\)

           \(=>x+1=1991\)

                  \(x\)       \(=1991-1\)

                  \(x\)       \(=1990\)

Vậy \(x=1990\)

1.

$2(x^2-4x+4)=(x-2)(x+3)$
$\Leftrightarrow 2(x-2)^2=(x-2)(x+3)$

$\Leftrightarrow 2(x-2)^2-(x-2)(x+3)=0$

$\Leftrightarrow (x-2)[2(x-2)-(x+3)]=0$

$\Leftrightarrow (x-2)(x-7)=0$

$\Leftrightarrow x-2=0$ hoặc $x-7=0$

$\Leftrightarrow x=2$ hoặc $x=7$

2.

$4x^2=9$

$\Leftrightarrow (2x)^2-3^2=0$

$\Leftrightarrow (2x-3)(2x+3)=0$

$\Leftrightarrow 2x-3=0$ hoặc $2x+3=0$

$\Leftrightarrow x=\frac{3}{2}$ hoặc $x=\frac{-3}{2}$
3.

$9x^2-1=(3x+1)(x+2)$

$\Leftrightarrow (3x+1)(3x-1)=(3x+1)(x+2)$

$\Leftrightarrow (3x+1)(3x-1)-(3x+1)(x+2)=0$

$\Leftrightarrow (3x+1)(3x-1-x-2)=0$

$\Leftrightarrow (3x+1)(2x-3)=0$

$\Leftrightarrow 3x+1=0$ hoặc $2x-3=0$

$\Leftrightarrow x=\frac{-1}{3}$ hoặc $x=\frac{3}{2}$

Các bạn giúp mình nha gợi ý : lưu ý trong 3 số tự nhiên liên tiếp có 1số chia hết cho 3

Bạn lưu ý, khi đăng đề thì đăng đầy đủ đề (bao gồm cả điều kiện và yêu cầu).

Đề yêu cầu tìm $x,y$?

$x,y$ là số như thế nào? Số nguyên? Số tự nhiên?

Bạn nên ghi rõ ra để mọi người hỗ trợ nhanh hơn nhé. 

\(x+2xy-2y=5\)

\(x+2y\times\left(x-1\right)=5\)

\(\left(x-1\right)+2y\times\left(x-1\right)=5-1\)

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

Ta có: 4 = (-1) x (-4) = (-2) x (-2) = 2 x 2 = 1 x 4

Ta lập bảng:

⇒ (x; y) ϵ {(5; 0); (-3; -1)}

\(\dfrac{3}{x}+\dfrac{4}{x-1}=\dfrac{5x-2}{x^2-x}\left(x\notin\left\{0;1\right\}\right)\\ \Leftrightarrow\dfrac{3\left(x-1\right)}{x\left(x-1\right)}+\dfrac{4x}{x\left(x-1\right)}=\dfrac{5x-1}{x\left(x-1\right)}\\ \Leftrightarrow3\left(x-1\right)+4x=5x-1\\ \Leftrightarrow3x-3+4x=5x-1\\ \Leftrightarrow7x-3=5x+1\\ \Leftrightarrow7x-5x=1+3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=\dfrac{4}{2}\\ \Leftrightarrow x=2\left(tm\right)\)

Xét đường tròn (O) có tiếp tuyến MB tại B nên 

\(\widehat{MBI}=\dfrac{1}{2}sđ\stackrel\frown{IB}\)

Lại có \(\widehat{IBH}=90^o-\widehat{BIH}\)

\(=90^o-\widehat{OIB}\)

\(=90^o-\dfrac{180^o-\widehat{IOB}}{2}\)

\(=\dfrac{180^o-180^o+sđ\stackrel\frown{IB}}{2}\)

\(=\dfrac{1}{2}sđ\stackrel\frown{IB}\)

Do đó \(\widehat{MBI}=\widehat{IBH}\) hay BI là tia phân giác của \(\widehat{MBH}\)

\(\Rightarrow d\left(I,MB\right)=d\left(I,BH\right)=IH=R_I\)

Suy ra MB là tiếp tuyến của (I)

Bài 4:

\(a)x-7< 2-x\\ \Leftrightarrow x+x< 2+7\\ \Leftrightarrow2x< 9\\ \Leftrightarrow x< \dfrac{9}{2}\\ b)x+2\le2+3x\\ \Leftrightarrow3x-x\ge2-2=0\\ \Leftrightarrow2x\ge0\\ \Leftrightarrow x\ge0\\ c)4+x>5-3x\\ \Leftrightarrow x+3x>5-4\\ \Leftrightarrow4x>1\\ \Leftrightarrow x>\dfrac{1}{4}\\ d)-x+7\ge x-3\\ \Leftrightarrow x+x\le7+3\\ \Leftrightarrow2x\le10\\ \Leftrightarrow x\le\dfrac{10}{2}=5\)

Lời giải:

$4\times 5+3=23$

$6\times 7+3=45$

Vậy $(*)$ là phép toán nhân hai số đã cho với nhau rồi cộng thêm 3.

Áp dụng vào thì $9(*)10=9\times 10+3=93$

Bài 2:

\(a)\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\\ \Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-4\left(x+1\right)=5\\ \Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\\ \Leftrightarrow-14x-9=5\\ \Leftrightarrow-14x=9+5=14\\ \Leftrightarrow x=\dfrac{14}{-14}\\ \Leftrightarrow x=-1\\ b)\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\\ \Leftrightarrow\left(25x^2+10x+1\right)-\left(25x^2-9\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow10x+10=30\\ \Leftrightarrow10x=30-10\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=\dfrac{20}{10}=2\)

Bài 1:

a: Sửa đề: \(A=6-2x+x^2\)

\(=x^2-2x+1+5\)

\(=\left(x-1\right)^2+5>=5\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(B=2x^2+3x-5\)

\(=2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{49}{8}>=-\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{4}=0\)

=>\(x=-\dfrac{3}{4}\)

c: \(C=4x^2+8x+1\)

\(=4x^2+8x+4-3\)

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

Dấu '=' xảy ra khi 2x+2=0

=>2x=-2

=>x=-1