TH 1 : \(x + 3/7 = 0\)

           \(x = 0 - 3/7\)

           \(x = -3/7\)

TH 2 : \(x-2/5=0\)

           \(x=0+2/5\)

            \(x=2/5\)

\(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)

\(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)

   ( 2x +1 ) 4 = 81

= ( 2x + 1 ) 4 = 34

= 2x + 1 = 3

= 2x = 2

= x = 1

HT

\(\left(2x-1\right)^4=81\)

Đặt \(\left(2x-1\right)^2=t;Đk:t\ge0\)

\(t^2=81\Rightarrow t=9\)

\(\left(2x-1\right)^2=9\Rightarrow2x-1=\pm3\)

\(2x-1=3\Rightarrow x=2\)

\(2x-1=-3\Rightarrow x=-1\)

\(x=\frac{4^2.4^3}{2^{10}}=\frac{4^5}{2^{10}}=\frac{\left(2^2\right)^5}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

TL
 

c2 = \(\frac{4^2.4^3}{2^{10}}\)=\(\frac{2^4.2^6}{2^{10}}\)= 1

HT

\(\frac{x}{3}\)=\(\frac{y}{4}\)=>x=\(\frac{3}{4}\)y

\(\frac{y}{3}\)=\(\frac{z}{5}\)=> z=\(\frac{5}{3}\)y

Ta có: 2x -3y+z=6

=>2.\(\frac{3}{4}\)y-3y+\(\frac{5}{3}\)y=6

=>\(\frac{1}{6}\)y=6

=> y=36

mà x= 3/4y=> x=27

z=5/3y=> z=60

x = 27

y = 36

z = 60

Theo giả thiết ta có: 

Khi đó ta có: 

\(\frac{x}{9}\)= \(\frac{y}{12}\)=\(\frac{z}{20}\)=\(\frac{2x}{18}\)= \(\frac{3y}{36}\)= \(\frac{z}{20}\)= \(\frac{2x-3y+z}{18-36+20}\)=\(\frac{6}{2}\)= 3 ( tính chất tỉ lệ thức )

Vậy x = 3.9 = 27; y = 3.12 = 36 ; z = 3.20 = 60

HT

\(\frac{x}{y}\)=\(\frac{7}{13}\)=> x=\(\frac{7}{13}\)y

Ta có : -x=y-40

=> x+y=40

Thay x =\(\frac{7}{13}\)y

=> \(\frac{7}{13}\)y+y=40

=>\(\frac{20}{13}\)y=40

=> y= 26

Vậy y =26

x=\(\frac{7}{13}\)y=> x=14

\(x=-y+40\)

\(\frac{x}{y}=\frac{7}{13}\Rightarrow\frac{-y+40}{y}=\frac{7}{13}\Rightarrow-13y+520=7y\)

\(\Rightarrow20y=520\Rightarrow y=260\Rightarrow x=-220\)

a/ \(\left(3x-1\right)^6=\left(3x-1\right)^4\Rightarrow\left(3x-1\right)=\left\{-1;0;1\right\}\)

\(\Rightarrow x=\left\{0;\frac{1}{3};\frac{2}{3}\right\}\)

b/

\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b-c+a-b+c-a+b+c}{a+b+c}=1\)

\(\Rightarrow\frac{a+b-c}{c}=1\Rightarrow a+b=2c\)

Tương tự

\(b+c=2a;a+c=2b\)

\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=8\)

=0 nha !

nhớ tích cho mìn nhe!

P=|x−2016|+|x−2018|+|x−2020|P=|x-2016|+|x-2018|+|x-2020|

⇔P=|x−2016|+|2020−x|+|x−2018|⇔P=|x-2016|+|2020-x|+|x-2018|

⇔P≥|x−2016+2020−x|+0⇔P≥|x-2016+2020-x|+0

⇔P≥4036⇔P≥4036

Vậy Pmin=4036Pmin=4036 khi {(x−2016)(2020−x)≥0x−2018=0{(x−2016)(2020−x)≥0x−2018=0

                                    ⇔x=2018

em ko biết em mới lớp 5 nha

Giải thích các bước giải: Ta có : a+c=2b, 2bd=c(b+d)

-> 2bd=(a+c)d =c(b+d)

-> ad+cd = bc+cd

-> ad=bc

-> a/b=c/d

ui cảm ơn bạn

TL:

Không nhìn thấy hình

_HT_