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Gọi số hoa ba bạn An, Hồng và Liên hái được lần lượt là \(x;y;z\) và \(x;y;z\) tỉ lệ với \(4;5;6\left(x;y;z\in N\cdot\right)\)
Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}\) và \(x+y+z=75\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y+z}{4+5+6}=\dfrac{75}{15}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.4=20\\y=5.5=25\\z=5.6=30\end{matrix}\right.\)
Vậy số hoa mà ba bạn An, Hồng và Liên hái được lần lượt là 20 bông hoa; 25 bông hoa; 30 bông hoa.
Gọi số hoa An,Hồng,Liên hái được lần lượt là a,b,c
Theo đề, ta có: a/4=b/5=c/6
Áp dụng tính chât của DTSBN, ta được:
\(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{c}{6}=\dfrac{a+b+c}{4+5+6}=\dfrac{75}{15}=5\)
=>a=20; b=25; c=30
Gọi số hoa An, Bình,Công lần lượt là a,b,c ta có:
\(\frac{a}{4}\)=\(\frac{b}{5}\)=\(\frac{c}{6}\)
ta có: \(\frac{c-a}{6-4}\)=\(\frac{18}{2}\)=9
do đó:\(\frac{a}{4}\)=9\(\Rightarrow\)a=36
\(\frac{b}{5}\)=9\(\Rightarrow\)b=45
\(\frac{c}{6}\)=9\(\Rightarrow\)c=54
Vậy số hoa của An,Bình,Công lần lượt là 36;45;54
Goi a, b, c (bong hoa) lan luot la so bong hoa cua moi ban Lan, Ha, Hai
Theo de ta co:\(\frac{a}{4}=\frac{b}{3}=\frac{c}{5}\)va a+b+c=48
Ta co: \(\frac{a}{4}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{4+3+5}=\frac{48}{12}=4\)
\(\frac{a}{4}=4\Rightarrow a=4\cdot4=16\)
\(\frac{b}{3}=4\Rightarrow b=3\cdot4=12\)
\(\frac{c}{5}=4\Rightarrow c=5\cdot4=20\)
Vay so bong hoa cua ba ban Lan, Ha, Hai da gap lan luot la 16 bong, 12 bong, 20 bong.
Áp dụng t/c của dãy tie số bằng nhau, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y+z}{4+5+6}=\dfrac{75}{15}=5\)
Khi đó:
\(\dfrac{x}{4}=5\Rightarrow x=5.4=20\left(bông\right)\)
\(\dfrac{y}{5}=5\Rightarrow y=5.5=25\left(bông\right)\)
\(\dfrac{z}{6}=5\Rightarrow z=5.6=30\left(bông\right)\)
Giải thích các bước giải:
Gọi số hoa điểm tốt của ba bạn An,Bình,Cường lần lượt là:a,b,c( a,b,c >0)
Theo đề bài ta có: \(\dfrac{a}{4}\)=\(\dfrac{b}{5}\)=\(\dfrac{c}{6}\) và a+b+c=75
Áp dụng t/chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{4}\)=\(\dfrac{b}{5}\)=\(\dfrac{c}{6}\)=\(\dfrac{a+b+c}{4+5+6}\)=\(\dfrac{75}{15}\)=5
=>\(\dfrac{a}{4}\)=5=>a=20
\(\dfrac{b}{5}\)=5=>b=25
\(\dfrac{c}{6}\)=5=>c=30
Vậy số hoa điểm tốt của ba bạn An,Bình,Cường lần lượt là: 20;25;30 ( bông hoa)
Gọi số hoa điểm tốt của ba bạn An,Bình,Cường lần lượt là:a,b,c( a,b,c >0)
Theo đề bài ta có: a/4=b/5=c/6 và a+b+c=75
Áp dụng t/chất của dãy tỉ số bằng nhau ta có:
a/4=b/5=c/6=a+b+c/4+5+6=75/15=5
=>a/4=5=>a=20
b/5=5=>b=25
c/6=5=>c=30
Vậy số hoa điểm tốt của ba bạn An,Bình,Cường lần lượt là: 20;25;30 ( điểm
Gọi số hoa đạt được của An, Bình, Cường lần lượt là a ; b ; c ( a ; b ; c > 0 )
Theo bài ra ta có \(\frac{a}{4}=\frac{b}{5}=\frac{c}{6};a+b+c=75\)
Theo tc dãy tỉ số bằng nhau
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\frac{a+b+c}{4+5+6}=\frac{75}{15}=5\Rightarrow a=20;b=25;c=30\)(tm)
Vậy ...
Gọi số hoa của 3 bạn lần lượt là x, y , z
Vì x,y,z TLT vớ 4,5,6
=> x/4=y/5=z/6=k
Theo t/c dãy tỉ số bằng nhau :
k= x+y+z/ 4+5+6 = 75/15=5
=> x= 5.4=20
y= 5. 5 = 25
z= 5.6=30
Vậy ..
Gọi số hoa 3 bạn hái được lần lượt là a,b,c \(\left(a,b,c\inℕ^∗\right)\)
Theo đề bài ra,ta có :
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\frac{a+b+c}{4+5+6}=\frac{75}{15}=6\)
\(\Rightarrow\hept{\begin{cases}a=6.4=24\\b=6.5=30\\c=6.6=36\end{cases}}\)
Vậy ....
Gọi 3 tổ có số học sinh lần lượt là: a, b, c ( \(a,b,c\inℕ^∗;a,b,c< 135\))
Theo bài ra, ta có: \(a+b+c=135\)và \(a:b:c=4:5:6\)
Từ \(a:b:c=4:5:6\)\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\frac{a+b+c}{4+5+6}=\frac{135}{15}=9\)
\(\Rightarrow a=9.4=36\); \(b=9.5=45\); \(c=9.6=54\)
Vậy số bông hoa của mỗi tổ lần lượt là 36 , 45 , 54
Gọi số hoa tuyết 3 lớp 7A,7B,7C làm lần lượt là a,b,c(hoa)
Áp dụng tích chất dãy tỉ số bằng nhau:
\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{b+c-a}{5+4-3}=\dfrac{84}{6}=14\)
\(\Rightarrow\left\{{}\begin{matrix}a=14.3=42\\b=14.4=56\\c=14.5=70\end{matrix}\right.\)
Vậy số hoa tuyết lớp 7A,7B,7C làm lần lượt là 42 bông, 56 bông, 70 bông
Gọi số hoa tuyết 7A;7B và 7C làm được lần lượt là x;y;z (hoa); x;y; z thuộc N*
Số hoa tuyết 7A làm ít hơn tổng số của 7B và 7C là 84 bông=>y+z-x=84
Số hoa tuyết 7A;7B và 7C làm tỉ lệ với 3;4;5 nên ta có: format('truetype')%3Bfont-weight%3Anormal%3Bfont-style%3Anormal%3B%7D%3C%2Fstyle%3E%3C%2Fdefs%3E%3Cline%20stroke%3D%22%23000000%22%20stroke-linecap%3D%22square%22%20stroke-width%3D%221%22%20x1%3D%222.5%22%20x2%3D%2214.5%22%20y1%3D%2220.5%22%20y2%3D%2220.5%22%2F%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20font-style%3D%22italic%22%20text-anchor%3D%22middle%22%20x%3D%228.5%22%20y%3D%2215%22%3Ex%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%228.5%22%20y%3D%2237%22%3E3%3C%2Ftext%3E%3Ctext%20font-family%3D%22math17f39f8317fbdb1988ef4c628eb%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2225.5%22%20y%3D%2226%22%3E%3D%3C%2Ftext%3E%3Cline%20stroke%3D%22%23000000%22%20stroke-linecap%3D%22square%22%20stroke-width%3D%221%22%20x1%3D%2236.5%22%20x2%3D%2248.5%22%20y1%3D%2220.5%22%20y2%3D%2220.5%22%2F%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20font-style%3D%22italic%22%20text-anchor%3D%22middle%22%20x%3D%2242.5%22%20y%3D%2215%22%3Ey%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2242.5%22%20y%3D%2237%22%3E4%3C%2Ftext%3E%3Ctext%20font-family%3D%22math17f39f8317fbdb1988ef4c628eb%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2259.5%22%20y%3D%2226%22%3E%3D%3C%2Ftext%3E%3Cline%20stroke%3D%22%23000000%22%20stroke-linecap%3D%22square%22%20stroke-width%3D%221%22%20x1%3D%2270.5%22%20x2%3D%2282.5%22%20y1%3D%2220.5%22%20y2%3D%2220.5%22%2F%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20font-style%3D%22italic%22%20text-anchor%3D%22middle%22%20x%3D%2276.5%22%20y%3D%2215%22%3Ez%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2276.5%22%20y%3D%2237%22%3E5%3C%2Ftext%3E%3C%2Fsvg%3E)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
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vậy số hoa tuyết 7A; 7B và 7C làm là: 42 hoa; 56 hoa và 70 hoa
Gọi số hoa của An , Hồng , Liên lần lượt là x , y , z( x , y , z > 0 )
ta có : \(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\) và \(x+y+z=75\)
Áp dụng t/c DTSBN :
\(\Rightarrow\frac{x}{4}=\frac{y}{5}=\frac{z}{6}=\frac{x+y+z}{4+5+6}=\frac{75}{15}=5\)
Số hoa của An là : \(\frac{x}{4}=5\Rightarrow x=20\)(bông )
Số hoa của Hồng là : \(\frac{y}{5}=5\Rightarrow y=25\)(bông )
Số hoa của Liên là: \(\frac{z}{6}=5\Rightarrow z=30\)( bông )
Đ/s...
\(=\frac{9}{4}+\frac{1}{4}:\left(-\frac{3}{4}\right)\)
\(=\frac{10}{4}.\left(-\frac{4}{3}\right)\)
\(=-\frac{10}{3}\)
Gọi số bông hoa ba bạn Hải, Cường, Oanh lần lượt cắt được là : x, y, z ( bông hoa ) ( ĐK : x, y, z thuộc N* )
Theo bài ra : x/4 = y/5 = z/6 và x+y+z=135
Áp dụng TC dãy tỉ số bằng nhau :
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z.}{6}=\dfrac{x+y+z}{4+5+6}=\dfrac{135}{15}=9\)
\(=>\left\{{}\begin{matrix}x=9.4=36\\y=9.5=45\\z=9.6=54\end{matrix}\right.\) (TM)
Vậy số hoa ba bạn cắt được lần lượt là : 36, 45, 54 bông