có j thắc mắc thì mn cứ hỏi ạ, em cần trc sáng mai nhé!? ><

b: Xét ΔABD và ΔBAC có

BA chung

BD=AC

AD=BC

Do đó: ΔABD=ΔBAC

c: ta có: EA+EC=AC

EB+ED=BD

mà AC=BD

và EA=EB

nên EC=ED

dấu / / là giá trị tuyệt đối

a) / x + 3 / = 15

=> x + 3 = 15 hay x + 3 = -15

TH1 : x + 3 = 15

           x =15 - 3

           x = 12

TH2 :  x + 3 = -5

             x = -5 - 3

             x = -8

VẬy : x thuộc { 12;-8 }

b ) /x-7/ + 13 = 25

      / x - 7 / = 25 - 13

      / x - 7 / = 12

   => x - 7 = 12 hay x -  7 = -12

TH1 : x - 7 = 12

         x = 12 + 7

         x =19

TH2: x - 7 = -12

          x = -12 + 7

          x = -5

Vậy : x thuộc { 19; -5 }

c) 3 ( x + 3 )² = 27

  ( x + 3 )² = 27 : 3

  ( x + 3 ) ² = 9

x + 3 = 3 hay x + 3 = -3 ( lũy thừa bậc chẵn )

Th1 : x + 3 = 3

         x = 3 -3 

         x = 0

TH2 : x + 3 = -3

           x = -3-3

       x = -6

Vậy : x thuộc { 0 ; -6 }

d) 20 - 2 ( 1-x )³ = 4

 2 ( 1 - x )³ = 20 - 4

 2 ( 1 - x )³ = 16

( 1 - x )³ = 16 : 2

 ( 1 -x )³ = 8

 ( 1 - x )³ = 2³ 

=> 1 - x = 2

   x = 1 - 2

   x = -1

VẬy : x = -1

C3:B

2B

3B

B

B

b, Áp dụng t/c dtsbn:

\(x:y:z=2:5:7\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x-y+z}{2-5+7}=\dfrac{25}{4}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{25}{2}\\y=\dfrac{125}{4}\\z=\dfrac{175}{4}\end{matrix}\right.\)

c, Áp dụng t/c dstbn:

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\\ \Rightarrow\left\{{}\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)

d, Áp dụng t/c dstbn:

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-2z}{8-12-2\cdot15}=\dfrac{36}{-34}=-\dfrac{18}{17}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{144}{17}\\y=-\dfrac{216}{17}\\z=-\dfrac{270}{17}\end{matrix}\right.\)

e, Áp dụng t/c dtsbn:

\(x:y:z=3:5:\left(-2\right)\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x-y+3z}{3\cdot5-5+3\left(-2\right)}=\dfrac{12}{4}=3\\ \Rightarrow\left\{{}\begin{matrix}x=9\\y=15\\z=-6\end{matrix}\right.\)

Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho!

tk mk đi mk tk bn rồi bn k mk di

vớ vẩn vãi tk cái đầu nhà m

a) \(A=\left(2\sqrt{12}-\sqrt{75}+\dfrac{1}{2}\sqrt{48}\right):\sqrt{3}\)

\(A=\left(4\sqrt{3}-5\sqrt{3}+2\sqrt{3}\right):\sqrt{3}\)

\(A=\sqrt{3}:\sqrt{3}\)

\(A=1\)

b) \(B=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(B=\left|2-\sqrt{5}\right|-\left|\sqrt{5}+1\right|\)

\(B=-2+\sqrt{5}-\sqrt{5}-1\)

\(B=-3\)

c) \(C=\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{3+\sqrt{7}}\)

\(C=\dfrac{3\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{4\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)

\(C=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\dfrac{4\left(3-\sqrt{7}\right)}{2}\)

\(C=\sqrt{7}+2-2\left(3-\sqrt{7}\right)\)

\(C=\sqrt{7}+2-6+2\sqrt{7}\)

\(C=3\sqrt{7}-4\)

d) \(D=3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\)

\(D=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)

\(D=5\sqrt{2a}-3a\sqrt{2a}-2\sqrt{2a}\)

\(D=3\sqrt{2a}-3a\sqrt{2a}\)

e) \(E=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)

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

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

\(E=\left(\sqrt{3}+1\right)-\left(\sqrt{3}+1\right)\)

\(E=0\)

Lời giải:

a. 

\(A=2\sqrt{\frac{12}{3}}-\sqrt{\frac{75}{3}}+\frac{1}{2}\sqrt{\frac{48}{3}}=2\sqrt{4}-\sqrt{25}+\frac{1}{2}\sqrt{16}\)

\(2.2-5+\frac{1}{2}.4=1\)

b. 

\(B=|2-\sqrt{5}|-|\sqrt{5}+1|=\sqrt{5}-2-(\sqrt{5}+1)=-3\)

c. 

\(C=\frac{3(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}-\frac{4(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}\)

\(=\frac{3(\sqrt{7}+2)}{7-2^2}-\frac{4(3-\sqrt{7})}{3^2-7}\)

\(=\frac{3(\sqrt{7}+2)}{3}-\frac{4(3-\sqrt{7})}{2}=\sqrt{7}+2-2(3-\sqrt{7})=-4+3\sqrt{7}\)

e. 

\(E=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}-\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\sqrt{3}+1-\frac{2(\sqrt{3}+1)}{3-1^2}=(\sqrt{3}+1)-(\sqrt{3}+1)=0\)

Những câu đã đăng rồi thì em hạn chế đăng lại nhé.

a: \(=\dfrac{\left(4\sqrt{3}-5\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}\right)}{\sqrt{3}}\)

=4-5+1/2*4

=-1+2

=1

b: \(=\left|2-\sqrt{5}\right|-\left|\sqrt{5}+1\right|\)

\(=\sqrt{5}-2-\sqrt{5}-1=-3\)

c: \(=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\dfrac{4\left(3-\sqrt{7}\right)}{2}\)

\(=\sqrt{7}+2-2\left(3-\sqrt{7}\right)\)

\(=\sqrt{7}+2-6+2\sqrt{7}=3\sqrt{7}-4\)

d: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)

\(=3\sqrt{2a}-3a\cdot\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}\)

\(=\sqrt{2a}\left(3-3a\right)\)

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

\(=\sqrt{3}-\sqrt{3}-1=-1\)