4/21

D 4/21

\(6\cdot x+9=15\)

\(\Rightarrow6\cdot x=15-9\)

\(\Rightarrow6\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{6}=1\)

_______________

\(43+2x=49\)

\(\Rightarrow2x=49-43\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=\dfrac{6}{2}=3\)

____________

\(x:2+5=11\)

\(\Rightarrow x:2=11-5\)

\(\Rightarrow x:2=6\)

\(\Rightarrow x=6\cdot2=12\)

______________

\(77-11x=0\)

\(\Rightarrow11x=77\)

\(\Rightarrow x=\dfrac{77}{11}\)

\(\Rightarrow x=7\)

_______________

\(12-4:x=8\)

\(\Rightarrow4:x=12-8\)

\(\Rightarrow4:x=4\)

\(\Rightarrow x=\dfrac{4}{4}=1\)

_____________

\(x:3+8=11\)

\(\Rightarrow x:3=11-8\)

\(\Rightarrow x:3=3\)

\(\Rightarrow x=3\cdot3=9\)

a: 6x+9=15

=>6x=6

=>x=1

b: 2x+43=49

=>2x=6

=>x=3

c: x:2+5=11

=>x:2=6

=>x=12

d: 77-11x=0

=>7-x=0

=>x=7

e: 12-4:x=8

=>4:x=4

=>x=1

f: x:3+8=11

=>x:3=3

=>x=9

a.x=1

b.x=3

c.x=12

d.x=7

e.x=1

f.x=9

\(\left(x+7\right)\left(3x-15\right)=0\\ \Rightarrow3\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\\ 4x\left(x+7\right)=2\left(x+7\right)\\ \Rightarrow4x\left(x+7\right)-2\left(x+7\right)=0\\ \Rightarrow2\left(2x-1\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-7\end{matrix}\right.\\ \left(x-3\right)^2-x\left(x-4\right)=5\\ \Rightarrow x^2-6x+9-x^2+4x-5=0\\ \Rightarrow-2x+4=0\\ \Rightarrow-2x=-4\Rightarrow x=2\)

hưng phúc                                                          đầy đủ chưa bạn nhỉ?

1) \(\left(x+7\right)\left(3x-15\right)=0\)

⇔\(\left[{}\begin{matrix}x+7=0\\3x-15=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

2) \(4x\left(x+7\right)=2\left(x+7\right)\)

\(2\left(2x+1\right)\left(x+7\right)=0\)

⇔\(\left[{}\begin{matrix}2x+1=0\\x+7=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-7\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-x\left(x-4\right)=5\)

\(\Leftrightarrow x^2-6x+9-x^2+4x=5\)

\(\Leftrightarrow-2x=-4\)

hay x=2

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

a: =>7/9:x=1/18-2/9=1/18-4/18=-3/18=-1/6

=>x=-7/9:1/6=-7/9*6=-42/9=-14/3

b: =>x*7/5=2/15+2/5=8/15

=>x=8/15:7/5=8/21

c: =>x-1/2=3/14:4/7=3/8

=>x=3/8+4/8=7/8

d: =>0,4x+0,3x-0,2x=0,7

=>0,5x=0,7

=>x=1,4

7/10

MG42E3RSDáidsdPDIOHSA8DR0RYE9FHEUCnd09f7eqg03

\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)