cần giúp gấp vs ạ huhu ( đề bài : tìm x biêt)
cần giúp gấp vs ạ huhu ( đề bài : tìm x biêt)
cần giúp gấp vs ạ, cứu em với (đề bài: tìm x biết)
a: \(\left(x+10\right)\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)
b: \(\left(2x+10\right)\left(4+x\right)=0\)
=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)
c: \(\left(4x+20\right)\left(12x-24\right)=0\)
=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d: \(\left(x-2024\right)\left(4x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)
e: \(\left(2x-6\right)\left(7+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
g: (4x+8)(6-x)=0
=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
h: (2x+2)(4x-8)=0
=>2(x+1)*4*(x-2)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
i: (2x-2024)(8x-16)=0
=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)
=>\(\left(x-1012\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)
Giúp mình vs ạ, cần gấp huhu
Bài có khúc bị khuyết em nha! Mà lại khúc quan trọng nữa
Các bạn giúp mình giải bài 4 hình vói ạ chỉ cần 3 câu đầu là đc ah Cứu mk vs mk đang gấp huhu
Giúp mình vs ah cần gấp lém ạ huhu
It is believed that he won the prize in the contest yesterday.
He is believed to have won the prize in the contest yesterday.
Giúp mình câu 4 vs ạ, cần gấp huhu
a) \(M_X=M_{Br2}=160\) (đvC)
b) CT của hợp chất : X2O3
Ta có : \(2X+16.3=160\)
=> X=56
Vậy X là Fe
Mọi ng giải hộ mik câu này vs ạ!
Tìm x bài phân tích đa thức thành nhân tử bằng phương pháp nhóm các hạng tử
Đề bài :x^3 -4x +x -2 =0
Mọi ng giúp mik vs ạ mik đng cần gấp mik cảm ơn mọi ng nhìều!!
đề bài : chuyển sang câu bị động giúp mk vs với ạ , mk cần râts gấp ạ
8 Her telephone number isn't known by me
9 The children will be brought home by my students
10 đúng r
11 We were given more information by her
12 All the workers of the plan were being instructed by the chief engineer
Her telephone number isn't known by me
The children will be brought home by my students
A present will be sent to me last week
we were gave more information by her
all the workers of the plan were being instrcued by the chief engineer
8. Her telephone number isn't know by me.
9. The children will be brought home by my students.
10. A present was sent me last week by them.
11. We were given more information by her.
12. All the workers of the plan were being instructed by the chief engineer.
Các bạn giúp mik nhanh vs ạ! Huhu mik đang cần gấp lắm
\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)
\(=\dfrac{x^3-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\)
\(=\dfrac{x^3-x^2-2x+2x-4}{x^2-4}\)
\(=\dfrac{x^3-x^2-4}{x^2-4}\)
a,\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=x-1\)
b, C=0\(\Rightarrow x-1=0\Rightarrow x=1\)
c, Để C nhận giá trị dương thì \(x-1\ge0\Rightarrow x\ge1\)