2x=3y

=>\(\dfrac{x}{3}=\dfrac{y}{2}\)

=>\(\dfrac{x}{9}=\dfrac{y}{6}\)

4y=3z

=>\(\dfrac{y}{3}=\dfrac{z}{4}\)

=>\(\dfrac{y}{6}=\dfrac{z}{8}\)

=>\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

mà x-y+2z=57

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x-y+2z}{9-6+2\cdot8}=\dfrac{57}{19}=3\)

=>x=27; y=18; z=24

mk ko ghi dau dc

\(\frac{2020-4x}{x}=100x\)

<=> 2020 - 4x = x.100x

<=> 2020 - 4x = 100x²

<=> 100x² + 4x - 2020 = 0 

<=> 4( 25x² + x - 505 ) = 0

<=> 25x² + x - 505 = 0

Tới đây không giải nữa :)) Lớp 6 làm gì đã học pt bậc 2 :)) 

Xem lại đề nhé ^^

\(\frac{\left(2020-4x\right)}{x}=100\)

đề như này à

\(\text{a) 3.(x-2)+x.(x-2)=0}\)
\(\Leftrightarrow\)\(\text{(x-2)(3+x)=0}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\3+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-3}\)
\(b,4x.\left(x-2\right)-x+2\)=0
\(\Leftrightarrow4x.\left(x-2\right)-\left(x-2\right)\)=0
\(\Leftrightarrow\left(x-2\right)\left(4x-1\right)\)=0
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{4}\end{array}\right.\)
Vậy x=2 hoặc \(x=\frac{1}{4}\)
 

a) 3.(x-2) + x. ( x-2) = 0

(x - 2)(3 + x) = 0

TH1:

x - 2 = 0

x = 2

TH2:

3 + x = 0

x = -3

Vậy x = 2 hoặc x = -3

b) 4x.(x-2) -x +2 = 0

4x(x - 2) - x + 2 = 0

(x - 2)(4x - 1) = 0

TH1:

x - 2 = 0

x = 2

4x - 1 = 0

4x = 1

x = 1/4

Vậy x = 2 hoặc x = 1/4

a) 3 . ( x - 2 ) + x . ( x - 2 ) = 0

=> 3x - 6 + 2x - 2x = 0

=> 3x + 2x - 2x = 0 + 6

=> 3x = 6

=> x = 6 : 3 = 2

b) 4x . ( x - 2 ) - x + 2 = 0

=> 5x - 6x - x + 2 = 0

=> 5x - 6x - x = 0 - 2 = - 2

=> - 2x = - 2

=> x = - 2 : ( - 2 )

=> x = 1

\(\left|x+5\right|+\left(3y-4\right)^{2010}=0\)

Vì \(\left|x+5\right|\ge0\forall x\)

Vì \(\left(3y-4\right)^{2010}\ge0\forall y\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2010}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)

a) pt

<=> (x - 5)(x + 5) - (x - 5) = 0

<=> (x - 5)(x + 4) = 0

<=> x - 5 = 0 hoặc x + 4 = 0

<=> x = 5 hoặc x = -4

b) pt

<=> (2x - 1)(2x - 1 - 2x - 1) = 0

<=> (2x - 1).(-2)=0

<=> 2x - 1 = 0

<=> x = 1/2

c) pt

<=> (x - 1)(x + 1)(x^2 + 4) = 0

<=> x - 1 = 0 hoặc x + 1 = 0 hoặc x^2 + 4 = 0

<=> x = 1 hoặc x = -1

a,x2−52−(x−5)=0<=>(x−5)(x+5)−(x−5)=0<=>(x−5)(x+4)=0=>x=5;x=−4.b,x2−x−6=0<=>x2−3x+2x−6=0<=>x(x−3)+2(x−3)=0<=>(x+2)(x−3)=0=>x=3;x=−2

a. x² - 25 - (x - 5) = 0

<=> x² - 5² - (x - 5) = 0

<=> (x - 5)(x + 5) - (x - 5) = 0

<=> (x + 5 - 1)(x - 5) = 0

<=> (x + 4)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x+4=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)

b. (2x - 1)² - (4x² - 1) = 0

<=> (2x - 1)² - (2x - 1)(2x + 1) = 0

<=> (2x - 1)(1 - 2x + 1) = 0

<=> (2x - 1)(2 - 2x) = 0

<=> \(\left[{}\begin{matrix}2x-1=0\\2-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

c. x²(x² + 4) - x² - 4 = 0

<=> x²(x² + 4) - (x² + 4) = 0

<=> (x² - 1)(x² + 4) = 0

<=> (x - 1)(x + 1)(x² + 4) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x+1=0\\x^2+4=0\left(VLí\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

B = 4x - x²

B = -(x² - 4x)

B = -(x² - 2.2x + 4 - 4)

B = -(x - 2)² + 4

Vi -(x - 2)² <= 0 voi moi x

=> -(x - 2)²  + 4 <= 4

Dau "=" xay ra <=> x - 2 = 0

                        <=> x       = 2

Vay GTLN cua B la 4 khi va chi khi x = 2

\(A^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\)

\(\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)

\(\Rightarrow-5\le A\le5\)

\(A_{max}=5\) khi \(x=y=1\)

\(A_{min}=-5\) khi \(x=y=-1\)