bạn nào giải thích giúp mik vs ạ. mik ko hỉu
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Tham khảo link : https://hoc24.vn/cau-hoi/bai-6-tim-n-thuoc-z-de-phan-so-a-dfrac20n-134n-3a-a-co-gia-tri-nho-nhat-b-a-co-gia-tri-nguyen.160524630905
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\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
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23:
u4=10 và u7=22
=>\(\left\{{}\begin{matrix}u1+3d=10\\u1+6d=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3d=-12\\u1+3d=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}d=4\\u1=10-12=-2\end{matrix}\right.\)
=>Chọn C
Câu 22:
\(\left\{{}\begin{matrix}u1+2u5=0\\S_4=14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}u1+2\left(u1+4d\right)=0\\4\cdot\dfrac{\left[2u1+3d\right]}{2}=14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3u1+8d=0\\2u1+3d=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6u1+16d=0\\6u1+9d=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7d=-21\\2u_1+3d=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=-3\\2u_1=7-3d=7+9=16\end{matrix}\right.\)
=>\(u_1=8;d=-3\)
=>Chọn A
21A
19B
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Theo NTBS ta có :
\(\begin{cases}
A+G = 0,5
\\G - A = 0,15
\end{cases}\)\\
=> A = T = 17,5%
G = X = 32,5%
Lại có A = \(\dfrac{A1+A2}{2} = \dfrac{A1+0,1}{2} = 0,175 \)
=> A1 = T2 = 25%
T1 = A2 = 10%
X1=G2 = 30%
G =\(\dfrac{G1+G2}{2} =\dfrac{G1+ 0,3}{2} = 0,325 \)
=> G1 = X2 = 35%
Theo bài ta có : \(\left\{{}\begin{matrix}\%A+\%G=50\%\\\%G-\%A=15\%\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\%A=\%T=17,5\%\\\%G=\%X=32,5\%\end{matrix}\right.\)
Mạch 1 có \(\%T_1=10\%=\%A_2\rightarrow\%A_1=\%T_2=2.\%A-\%T_1=25\%\)
\(\%X_1=30\%=\%G_2\rightarrow\%G_1=\%X_2=2.\%G-\%X_1=35\%\)
Vậy \(A_2=10\%,T_2=25\%,G_2=30\%,X_2=35\%\)
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1:
a: =x^2+3x+4x+12
=x(x+3)+4(x+3)
=(x+3)(x+4)
b: =4x^2-4x-5x+5
=4x(x-1)-5(x-1)
=(x-1)(4x-5)
c: =2x^2-3x-4x+6
=x(2x-3)-2(2x-3)
=(2x-3)(x-2)
3:
a: =2x^2-6xy+xy-3y^2
=2x(x-3y)+y(x-3y)
=(x-3y)(2x+y)
b: =x^2+3xy-xy-3y^2
=x(x+3y)-y(x+3y)
=(x+3y)*(x-y)
c: =6x^2+4xy-3xy-2y^2
=2x(3x+2y)-y(3x+2y)
=(3x+2y)(2x-y)
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\(E=22x-23-5x+2-3x+1\)
\(=14x-20\)
\(=14\cdot\dfrac{-2}{3}-20=\dfrac{-28}{3}-\dfrac{60}{3}=-\dfrac{88}{3}\)
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100 lít nước nở thêm: 100.27 = 2700cm3 = 2,7 lít
Thể tích nước trong bình: 100 + 2,7 = 102,7 lít
Chọn B
Gọi CTHH là: XH3
Theo đề, ta có: \(d_{\dfrac{XH_3}{H_2}}=\dfrac{M_{XH_3}}{M_{H_2}}=\dfrac{M_{XH_3}}{2}=8,5\left(lần\right)\)
=> \(M_{XH_3}=17\left(g\right)\)
Ta có: \(M_{XH_3}=M_X+1.3=17\left(g\right)\)
=> MX = 14(g)
Dựa vào bảng hóa trị, suy ra:
X là nitơ (N)
=> CTHH của hợp chất là NH3
Chọn B