Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
GIÚP MIK VS NHA:(((((
CẢM ƠN RẤT NHIỀU
MN XONG CÂU NÀO THÌ CỨ GỬI LUÔN CHO MIK CÂU ĐÓ NHA;-;
MIK CÒN CHÉP KỊP
:(((((((((((((( NHANHH NHANH GIÚP MIK Ạ
Câu 1:
\(a,\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{x-y}{4-7}=\dfrac{-15}{-3}=5\\ \Rightarrow\left\{{}\begin{matrix}x=20\\y=35\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{-32}{8}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-20\end{matrix}\right.\\ c,\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{-90}{10}=-9\\ \Rightarrow\left\{{}\begin{matrix}x=-18\\y=-27\\z=-45\end{matrix}\right.\\ d,\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x-4y+3z}{8-8+21}=\dfrac{42}{21}=2\\ \Rightarrow\left\{{}\begin{matrix}x=8\\y=4\\z=14\end{matrix}\right.\)
\(e,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{z-x}{7-5}=\dfrac{30}{2}=15\\ \Rightarrow\left\{{}\begin{matrix}x=75\\y=90\\z=105\end{matrix}\right.\\ f,\Rightarrow\dfrac{x}{3}=\dfrac{y}{5};\dfrac{x}{4}=\dfrac{z}{3}\Rightarrow\dfrac{x}{12}=\dfrac{y}{20}=\dfrac{z}{9}=\dfrac{x-y-z}{12-20-9}=\dfrac{-68}{-17}=4\\ \Rightarrow\left\{{}\begin{matrix}x=48\\y=80\\z=36\end{matrix}\right.\\ g,\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{6+4+3}=\dfrac{65}{13}=5\\ \Rightarrow\left\{{}\begin{matrix}x=30\\y=20\\z=15\end{matrix}\right.\\ h,\Rightarrow\dfrac{x}{4}=\dfrac{y}{6};\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{x}{20}=\dfrac{y}{30}=\dfrac{z}{48}=\dfrac{5x-3y-3z}{100-90-144}=\dfrac{-536}{-134}=4\\ \Rightarrow\left\{{}\begin{matrix}x=80\\y=120\\z=192\end{matrix}\right.\)
a: \(-6\cdot\left(-\dfrac{2}{3}\right)\cdot0.25=6\cdot\dfrac{2}{3}\cdot\dfrac{1}{4}=4\cdot\dfrac{1}{4}=1\)
b: \(\dfrac{-15}{4}\cdot\dfrac{-7}{15}\cdot\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{12}{5}\)
\(=\dfrac{84}{20}=\dfrac{21}{5}\)
c: \(\left(-2\dfrac{1}{5}\right)\cdot\left(-\dfrac{9}{11}\right)\cdot\left(-\dfrac{1}{14}\right)\cdot\dfrac{2}{5}\)
\(=-\dfrac{11}{5}\cdot\dfrac{2}{5}\cdot\dfrac{9}{11}\cdot\dfrac{1}{14}\)
\(=-\dfrac{11}{11}\cdot\dfrac{2}{14}\cdot\dfrac{9}{25}\)
\(=-\dfrac{9}{175}\)
\(a,=4\cdot0,25=1\\ b,=\dfrac{7}{4}\cdot\left(-\dfrac{12}{5}\right)=-\dfrac{21}{5}\\ c,=\left(-\dfrac{11}{5}\right)\left(-\dfrac{9}{11}\right)\left(-\dfrac{15}{14}\right)\cdot\dfrac{2}{5}\\ =\dfrac{9}{5}\cdot\left(-\dfrac{15}{14}\right)\cdot\dfrac{2}{5}=-\dfrac{27}{14}\cdot\dfrac{2}{5}=-\dfrac{27}{35}\\ d,=\left(-\dfrac{11}{2}\right)\left(-\dfrac{1}{2}\right)+\dfrac{4}{9}=\dfrac{11}{4}+\dfrac{4}{9}=\dfrac{115}{36}\\ e,=\dfrac{5}{4}\cdot\left(-\dfrac{8}{15}\right)-\dfrac{3}{5}-\dfrac{3}{10}=-\dfrac{2}{3}-\dfrac{3}{5}-\dfrac{3}{10}=-\dfrac{47}{30}\)
\(f,B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}=\dfrac{2\cdot6}{5\cdot3}=\dfrac{4}{5}\\ g,=\dfrac{5}{8}+\dfrac{9}{4}\cdot\dfrac{5}{3}-\dfrac{5}{24}=\dfrac{5}{8}+\dfrac{15}{4}-\dfrac{5}{24}=\dfrac{25}{6}\\ h,=\dfrac{49}{38}\cdot\left(\dfrac{152}{11}-\dfrac{57}{11}\right):\dfrac{245}{418}=\dfrac{49}{38}\cdot\dfrac{418}{245}\cdot\dfrac{95}{11}=\dfrac{95\cdot11}{5\cdot11}=19\\ k,=\dfrac{11}{30}+\dfrac{18}{35}\cdot\dfrac{35}{54}-\dfrac{18}{35}\cdot\dfrac{49}{18}-\dfrac{18}{35}\cdot\dfrac{28}{48}\\ =\dfrac{11}{30}+\dfrac{1}{3}-\dfrac{7}{5}-\dfrac{3}{10}=-1\)
https://hoc24.vn/cau-hoi/giai-het-giup-mk-vs-a-huhu.2030470567354
làm hết r mà :vv
ta có P(x)=x^2+ax+b ; Q(x)=x^2+cx+d
ta có x1 và x2 là nghiêm của P(x)Dán
nên \(x_1^2+ax_1+b=0;x_2^2+ax_2+b=0\)
\(\Rightarrow x_1^2=-ax_1-b\) và \(x_2^2=-ax_2-b\) (1)
Ta có x1,x2 là nghiêm của Q(x)
nên \(x_1^2+cx_1+d=0;x_2^2+cx_2+d=0\)
\(\Rightarrow x_1^2=-cx_1-d\)và \(x_2^2=-cx_2-d\) (2)
Từ (1) và (2) suy ra \(-ax_1-b=-cx_1-d\\ -ax_2-b=-cx_2-d\)
Do đó \(ax_1+b=cx_1+d\\ ax_2+b=+cx_2+d\)
Suy ra\(x_1^2+ax_1+b=x^2_1+cx_1+d\\ x^2_2+ax_2+b=x^2_2+cx_2+d\)
Nên P(x)=Q(x)
Q(x) =x2 +ax + b
P(x) = x2 +cx + d
Vì x1;x2 đều là nghiệm của P(x); Q(x)
=>x1;x2 là nghiệm của : P(x) - Q(x)=(c-a)x +(d-b)
=> PT: (c-a)x +(d-b) =0 có 2 nghiệm x1;x2
=>\(\left\{{}\begin{matrix}c-a=0\\d-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=c\\b=d\end{matrix}\right.\)
Nên => P(x) = Q(x) dpcm
Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)
Do đó: x=54; y=36
c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
Bài 3:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{a+b}{2+5}=\dfrac{70}{7}=10\)
Do đó: a=20; b=50