a) \(k=-5\)

b) 

Tìm x xong rồi tìm y

3 thì làm kiểu gì cũng được

Bài 3:

Hình 1:

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(x+60^0+40^0=180^0\)

=>\(x=180^0-100^0=80^0\)

Hình 2:

Xét ΔABD có \(\widehat{ADC}\) là góc ngoài tại đỉnh D

nên \(\widehat{ADC}=\widehat{ABD}+\widehat{BAD}=110^0\)

=>\(y=110^0\)

ΔDAC cân tại D

=>\(\widehat{DAC}=\widehat{DCA}=\dfrac{180^0-\widehat{ADC}}{2}\)

=>\(x=\dfrac{180^0-110^0}{2}=35^0\)

Hình 3:

Ta có: \(\widehat{CAB}+60^0=180^0\)

=>\(\widehat{CAB}=180^0-60^0=120^0\)

Xét ΔCAB có \(\widehat{CAB}+\widehat{ABC}+\widehat{ACB}=180^0\)

=>\(x+2x+120^0=180^0\)

=>\(3x=60^0\)

=>\(x=20^0\)

Hình 4:

Ta có: \(\widehat{ADC}+\widehat{ADB}=180^0\)

=>\(\widehat{ADB}=180^0-80^0=100^0\)

Xét ΔADB có

\(\widehat{ADB}+\widehat{ABD}+\widehat{BAD}=180^0\)

=>\(x+60^0+100^0=180^0\)

=>\(x=20^0\)

ta có: \(\widehat{ACD}+135^0=180^0\)

=>\(\widehat{ACD}=180^0-135^0=45^0\)

Xét ΔACD có \(\widehat{ACD}+\widehat{ADC}+\widehat{DAC}=180^0\)

=>\(y+80^0+45^0=180^0\)

=>\(y=55^0\)

Lời giải:

$4x-6=2x+4$

$\Leftrightarrow (4x-6)-(2x+4)=0$

$\Leftrightarrow 2x-10=0$

$\Leftrightarrow 2x=10$

$\Leftrightarrow x=5$

\(4x-6=2x+4\\ \Leftrightarrow4x-2x=4+6\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

Vậy phương trình có nghiệm là: \(x=5\)

what the hell sao nhìu zữ zậy?

cầu 24 :

a) ( 6.8⁴ - 5.8³ + 8² ) : 8²

= ( 6.4096 - 5.512 + 64 ) : 8²

= ( 24576 - 2560 + 64 ) : 8²

= ( 22016 + 64 ) : 8²

= 22080 : 8²

= 22080 : 64

= 345

b) ( 5.9² + 3⁵ - 2.3³ ) : 3²

= ( 5.81 + 243 - 2.27) : 3²

= ( 405 + 243 - 54 ) : 3²

= 594 : 3²

= 594 : 9

= 66

c) ( 2.3⁴ + 3² - 7.3³ ) : 3²

= ( 2.81 + 9 - 7.27 ) : 3²

= ( 162 + 9 - 189 ) : 3²

= -18 : 9

= -2

d) ( 6.2³ - 5.2⁴ + 2⁵ ) : 2³

= ( 6.8 - 5.16 + 32 ) : 2³

= ( 48 - 80 + 32 ) : 2³

= 0 : 2³

= 0 : 8 

= 0

câu 25 : mk hướng dẫn cách giải để nhanh hơn nhé !

a) thay x = -2 ; y = -2 ( hoặc x = y = -2 )vào biểu thức ... * bạn tự ghi lại đầu bài phép tính nhé ! " ta có :

rồi bạn thay phần x,y bằng số -2 vào, r sau đó bạn tính toán số mũ trc,  trong ngoặc tròn trc, ngoặc vuông rồi ngoặc nhọn và cuối cug là phép tính ở ngoài ngoặc nhé!

25:

a: \(=\dfrac{15x^5y^3}{5x^2y^2}-\dfrac{10x^3y^2}{5x^2y^2}+\dfrac{20x^4y^4}{5x^2y^2}\)

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

\(=3\cdot\left(-1\right)^3\cdot2-2\cdot\left(-1\right)+4\cdot\left(-1\right)^2\cdot2^2\)

\(=-6+2+16=18-6=12\)

b: \(=\dfrac{4x^4y^2+3x^4y^3-6x^3y^2}{x^2y^2}=4x^2+3x^2y-6x\)

\(=4\cdot\left(-2\right)^2+3\cdot\left(-2\right)^2\cdot\left(-2\right)-6\cdot\left(-2\right)\)

\(=4\cdot4-6\cdot4+6\cdot2=8-24+12=-4\)

c: \(=-3xy+6-9y^2=-3\cdot\dfrac{1}{2}\cdot4+6-9\cdot4^2\)

=-9*4^2

=-9*16=-144

d: \(=\dfrac{1}{6}y^3-\dfrac{1}{3}x^3=\dfrac{1}{6}\cdot3^3-\dfrac{1}{3}\cdot\left(-3\right)^3=\dfrac{191}{6}\)

Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)

         = - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)

         =  - \(\dfrac{5}{7}\)  + 2 + \(\dfrac{5}{7}\)

          = 2

2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8

   = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8

   =  \(\dfrac{2}{7}\) x 28 - 8

   = 8 - 8

    = 0 

3, \(\dfrac{37}{43}\) . \(\dfrac{17}{29}\) - \(\dfrac{21}{41}\).\(\dfrac{1}{2}\) + \(\dfrac{9}{58}\): 1\(\dfrac{6}{37}\) - \(\dfrac{6}{29}\): 1\(\dfrac{20}{21}\)

=    \(\dfrac{37}{43}\). \(\dfrac{17}{29}\) - \(\dfrac{21}{41}\) . \(\dfrac{1}{2}\) + \(\dfrac{9}{58}\).\(\dfrac{37}{43}\)  - \(\dfrac{6}{29}\).\(\dfrac{21}{41}\)

=    (\(\dfrac{37}{43}\).\(\dfrac{17}{29}\) + \(\dfrac{9}{58}\).\(\dfrac{37}{43}\)) - (\(\dfrac{21}{41}\).\(\dfrac{1}{2}\) + \(\dfrac{6}{29}\).\(\dfrac{21}{41}\))

=     \(\dfrac{37}{43}\).( \(\dfrac{17}{29}\) + \(\dfrac{9}{58}\)) - \(\dfrac{21}{41}\).( \(\dfrac{1}{2}\) + \(\dfrac{6}{29}\))

=       \(\dfrac{37}{43}\).\(\dfrac{43}{58}\) - \(\dfrac{21}{41}\).\(\dfrac{41}{58}\)

=         \(\dfrac{37}{58}\) - \(\dfrac{21}{58}\)

=          \(\dfrac{16}{58}\)

= \(\dfrac{8}{29}\)

11003 + 1677 + 1969 + 1031 + 1323 - 1003

= ( 11003 - 1003 ) + ( 1677 + 1323 ) + ( 1969 + 1031 )

= 10000 + 3000 + 3000

= 13000 + 3000

= 16000

\(=\left(\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}\right):\sqrt{\dfrac{5}{2}}\)

\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(=2\sqrt{5}.\dfrac{\sqrt{2}}{\sqrt{5}}=2\sqrt{2}\)

a) Ta có: \(\left(\dfrac{\sqrt{15}-\sqrt{20}}{\sqrt{3}-2}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{6}+\sqrt{5}}\right):\sqrt{\dfrac{5}{2}}\)

\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{10}}{2}\)

\(=2\sqrt{5}\cdot\dfrac{2}{\sqrt{10}}=2\sqrt{2}\)

1. B

2. C

3. C

4. B

5. A

6. D

7. C

8. C

9. A

10. B

1. B 2. C 3. C 4. B 5. A 6. D 7. C 8. C 9. A 10. B