Toán

=>11x-66=4x+11

=>7x=77

=>x=11

x = 11

x=11 nha

(a) Phương trình đường thẳng \(\left(d\right)\) có dạng tổng quát: \(y=ax+b\).

Do \(\left(d\right)\) đi qua \(A,B\) nên giá trị hoành độ và tung độ của \(A,B\) là các cặp nghiệm của phương trình đường thẳng.

\(\Rightarrow\left\{{}\begin{matrix}-3=a+b\\1=2a+b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=-7\end{matrix}\right.\).

Vậy: Phương trình đường thẳng \(\left(d\right):y=4x-7\).

(b)  Mình không hiểu rõ đề phần "có (1, 2)" ạ:D.

Sửa đề: cosx+sinx=1

=>\(\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)=1\)

=>\(sin\left(x+\dfrac{pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

=>\(\left[{}\begin{matrix}x+\dfrac{pi}{4}=\dfrac{pi}{4}+k2pi\\x+\dfrac{pi}{4}=\dfrac{3}{4}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2pi\\x=pi+k2pi\end{matrix}\right.\)

1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

3: A/B>3/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)

=>\(-\sqrt{x}+2>0\)

=>-căn x>-2

=>căn x<2

=>0<x<4

1) Thay x=64 vào A ta có:

\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)

2) \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

3) Ta có:

\(\dfrac{A}{B}>\dfrac{3}{2}\) khi

\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)

Mà: \(2\sqrt{x}\ge0\forall x\)

\(\Leftrightarrow2-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

Kết hợp với đk:

\(0< x< 4\)

1: ΔBED vuông tại E

=>DB^2=DE^2+EB^2

=>BE^2=DB^2-DE^2

ΔCED vuông tại E

=>CE^2+ED^2=CD^2

=>CE^2=CD^2-ED^2

BE^2-CE^2

=DB^2-DE^2-CD^2+DE^2

=DB^2-CD^2

2: DB^2-CD^2

=DB^2-AD^2(Do CD=AD)

=AB^2

mà DB^2-DC^2=BE^2-CE^2

nên BE^2-CE^2=AB^2

17:

ΔABC cân tại A

=>góc ABC=góc ACB=(180-góc BAC)/2

\(=\dfrac{180^0-120^0}{2}=30^0\)

Xét ΔABC có \(\dfrac{AB}{sinC}=\dfrac{BC}{sinA}\)

=>\(\dfrac{AB}{sin30}=\dfrac{48}{sin120}\)

=>\(AB=16\sqrt{3}\left(cm\right)\)

15:

Xét ΔABC có

\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\)

=>\(\dfrac{AC}{sin42}=\dfrac{20}{sin36}\)

=>\(AC\simeq22,77\left(cm\right)\)

16:

a:

ΔABC vuông tại A

=>BC^2=AB^2+AC^2

=>BC^2=15^2+20^2=625

=>BC=25cm

Xét ΔABC vuông tại A có AH là đường cao

nên AH*BC=AB*AC

=>AH*25=15*20=300

=>AH=12(cm)

b: BC=BH+CH=10cm

ΔABC vuông tại A có AH là đường cao

nên BH*BC=BA^2; CH*CB=CA^2

=>BA^2=2*10=20; CA^2=8*10=80

=>\(BA=2\sqrt{5}\left(cm\right);CA=4\sqrt{5}\left(cm\right)\)

Xét ΔABC vuông tại A có

\(cosB=sinC=\dfrac{AB}{BC}=\dfrac{2\sqrt{5}}{10}=\dfrac{\sqrt{5}}{2}\)

\(cosC=sinB=\dfrac{AC}{BC}=\dfrac{4\sqrt{5}}{10}=\dfrac{2\sqrt{5}}{5}\)

AF=FG=GC

=>AG=2/3AC

=>S ABG=2/3*S ABC

BE=ED=1/2AD

=>BE+ED=1/2AD+1/2AD=AD

=>BD=DA

=>D là trung điểm của AB

=>AD=1/2AB

=>BE=ED=1/4AB

AE=AD+DE=1/4AB+1/2AB=3/4AB

=>S AEG=3/4*S ABG=3/4*2/3*S ABC=1/2*S ABC

=>S BEGC=1/2*S ABC

AF=FG

=>AF=1/2*AG

=>S AEF=1/2*S AEG=1/2*1/2*S ABC=1/4*S ABC

AD=2DE

=>AD=2/3AE

=>S ADF=2/3*S AEF=2/3*1/4*S ABC=1/6*S ABC

=>s EDFG=S ABC(1/4-1/6)=1/12*S ABC

S BCGE-S DEFG=15cm²

=>S ABC(1/2-1/12)=15

=>S_ABC=15:5/12=36cm²

1: =>6x^2+21x-2x-7-6x^2+5x-6x+5=7

=>18x-2=7

=>18x=9

=>x=1/2

2: (3x+2)(2x+9)-(x+2)(6x+1)=7

=>6x^2+27x-4x-18-6x^2-x-12x-2=7

=>10x-20=7

=>10x=27

=>x=27/10

3: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

4: =>2(6x^2+15x-2x-5)-6(2x^2+4x-x-2)=-6

=>12x^2+26x-10-12x^2-18x+12=-6

=>8x+2=-6

=>8x=-8

=>x=-1

5: =>6x-2x^2-3+x+x^2+x-6=-(x^2-3x+2)

=>-x^2+8x-9+x^2-3x+2=0

=>5x-7=0

=>x=7/5

6: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

7: =>24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1-33

=>20x^2-16x-34=10x^2+3x-34

=>10x^2-19x=0

=>x(10x-19)=0

=>x=0 hoặc x=19/10

\(P=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{8\sqrt{x}}{x-4}\right):\dfrac{2\sqrt{x}+4-2\sqrt{x}-3}{\sqrt{x}+2}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}\cdot\dfrac{\sqrt{x}+2}{1}\)

\(=\dfrac{3x-6\sqrt{x}-x-2\sqrt{x}+8\sqrt{x}}{\sqrt{x}-2}=\dfrac{2x}{\sqrt{x}-2}\)

\(E=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)

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

\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\)

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

ΔABC vuông tại A

=>BC^2=AB^2+AC^2

=>\(BC=\sqrt{9^2+12^2}=15\left(cm\right)\)

Xét ΔABC có AD là phân giác

nên BD/AB=CD/AC

=>BD/3=CD/4

mà BD+CD+15

nên \(\dfrac{BD}{3}=\dfrac{CD}{4}=\dfrac{BD+CD}{3+4}=\dfrac{15}{7}\)

=>BD=45/7(cm)

Xét ΔABC có AD là phân giác

nên \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)

\(=\dfrac{2\cdot9\cdot12}{9+12}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{36\sqrt{2}}{7}\left(cm\right)\)

ΔABC vuông tại A có AK là đường cao

nên AK*BC=AB*AC

=>AK*15=12*9=108

=>AK=7,2cm

ΔAKD vuông tại K

=>AK^2+KD^2=AD^2

=>KD^2=AD^2-AK^2=1296/1225

=>KD=36/35(cm)