\(\left(x-\dfrac{1}{5}\right)\)² + (y + 0,4)¹⁰⁰ + (z - 3)⁶⁷⁸ = 0

Vì \(\left(x-\dfrac{1}{5}\right)^2\) ≥ 0; (y + 0,4)¹⁰⁰ ≥ 0; (z - 3)⁶⁷⁸ ≥ 0 ∀ \(x;y;z\) 

Vậy \(\left(x-\dfrac{1}{5}\right)^2\) + (y + 0,4)¹⁰⁰ + (z - 3)⁶⁷⁸ = 0 khi và chỉ khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\) ⇒  \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)

Vậy (\(x;y;z\)) = (\(\dfrac{1}{5}\); -0,4; 3)

\(\dfrac{y}{x}\) = \(\dfrac{7}{4}\) ⇒ \(\dfrac{y}{7}\) = \(\dfrac{x}{4}\) ⇒ (\(\dfrac{y}{7}\))² = (\(\dfrac{x}{4}\))² ⇒ \(\dfrac{y^2}{49}=\dfrac{x^2}{16}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

                   \(\dfrac{x^2}{16}\) = \(\dfrac{y^2}{49}\) = \(\dfrac{x^2-y^2}{16-49}\) = \(\dfrac{-33}{-33}\) = 1

                   \(x^2\) = 1.16 = 16 ⇒ \(x^2\) = 4² ⇒ \(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

                    y² = 49.1 = 49 ⇒ y² = 7² ⇒ \(\left[{}\begin{matrix}y=-7\\y=7\end{matrix}\right.\)

Vì \(\dfrac{y}{x}\) = \(\dfrac{7}{4}\) > 0 nên \(x;y\) cùng dấu

Vậy (\(x;y\)) = (-4; -7); (4; 7) 

\(\left(x-4\right)^2-3=1\\ \left(x-4\right)^2=1+3\\ \left(x-4\right)^2=4\\ \left(x-4\right)^2=\left(\pm2\right)^2\\ \left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\\ \left[{}\begin{matrix}x=2+4\\x=-2+4\end{matrix}\right.\\ \left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

(x - 4)^2 - 3 = 1

(x - 4)^2 = 1 + 3

(x- 4)^2 = 4

(x - 4)^2 = 2^2

x - 4 = 2

x = 2 + 4

x = 6

Vậy x = 6

a) Xét ΔBEA∆BEA và ΔCDA∆CDA có:

BA=CABA=CA (gt)

ˆAA^ chung

AE=ADAE=AD (gt)

⇒ΔBEA=ΔCDA⇒∆BEA=∆CDA (c.g.c)

⇒BE=CD⇒BE=CD (hai cạnh tương

A B C D E O

a) tam giác ABC có AB = AC (gt)

=> tam giác ABC cân tại A => góc B = góc C

lại có: D thuộc AB, E thuộc AC nên DB = AB - AD

                                                         EC = AC - AE

mà AB = AC, AD = AE => DB = EC

xét tam giác DBC và tam giác ECB có: DB = EC (cmt)

                                                              góc DBC = góc ECB (cmt)

                                                              BC: cạnh chung

=> tam giác DBC = tam giác ECB (cgc) => DC = BE (đpcm)

x+2/5=7/10

x=7/10-2/5

x=7/20-4/10

x=3/10

vậy x=3/10

3/10

x+1,(3)-1,(2)x=0,4

x+12/9-11/9x=0,4

x-11/9x=12/9-4/10

x-11/9x=120/90-36/90

x-11/9x=84/90

-2/9x=28/30

x=28/30:(-2/9)

x=28/30.-9/2

x=-252/60

x=-4,2

\(x+1,\left(3\right)\) - 1,\(\left(2\right)x\) = 0,4

\(x\) - 1,(2)\(x\) = 0,4 - (1,3)

\(x\) - \(\dfrac{11}{9}\)\(x\) =  \(\dfrac{2}{5}\) - \(\dfrac{12}{9}\)

    - \(\dfrac{2}{9}\)\(x\)     =   - \(\dfrac{14}{15}\)

         \(x=-\dfrac{14}{15}:\left(-\dfrac{2}{9}\right)\)

        \(x\) = \(\dfrac{21}{5}\)

Vậy \(x=\dfrac{21}{5}\)

Ta có: y(x+3)+x+14=0

=>y(x+3)+x+3+11=0

=>(x+3)(y+1)=-11

=>\(\left(x+3;y+1\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-2;-12\right);\left(-14;0\right);\left(-4;10\right);\left(8;-2\right)\right\}\)

Ta có: y(x+3)+x+14=0

=>y(x+3)+x+3+11=0

=>(x+3)(y+1)=-11

=>\(\left(x+3;y+1\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-2;-12\right);\left(-14;0\right);\left(-4;10\right);\left(8;-2\right)\right\}\)