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Bài 2:
a, Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{-5}=\dfrac{a+b}{2+\left(-5\right)}=\dfrac{21}{-3}=-7\)
(do \(a+b=21\))
\(\Rightarrow\left\{{}\begin{matrix}a=-7.2=-14\\b=-7.\left(-5\right)=35\end{matrix}\right.\)
Vậy \(a=-14;b=35\)
b, Áp dụng tính chất cảu dãy tỉ số bằng nhau ta có:
\(\dfrac{-10}{a}=\dfrac{-15}{b}=\dfrac{-10-\left(-15\right)}{a-b}=\dfrac{5}{-5}=-1\)
(do \(a-b=-5\))
\(\Rightarrow\left\{{}\begin{matrix}a=-10:\left(-1\right)=10\\b=-15:\left(-1\right)=15\end{matrix}\right.\)
Vậy \(a=10;b=15\)
Chúc bạn học tốt!!!
c, Ta có:
\(3x=2y\Rightarrow21x=14y\)
\(7y=5z\Rightarrow14y=10z\)
\(\Rightarrow21x=14y=10z\Rightarrow\dfrac{21x}{210}=\dfrac{14y}{210}=\dfrac{10z}{210}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\)
(do \(x-y+z=32\))
\(\Rightarrow\left\{{}\begin{matrix}x=2.10=20\\y=2.15=30\\z=2.21=42\end{matrix}\right.\)
Vậy \(x=20;y=30;z=42\)
Chúc bạn học tốt!!!
Câu 1
\(\left\{{}\begin{matrix}7A,7B\in N\\7B=7A+5\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}7B>7A\\\dfrac{7A}{7B}=\dfrac{8}{9}\end{matrix}\right.\)\(\dfrac{7A}{7B}=\dfrac{8}{9}\Rightarrow\dfrac{7A}{8}=\dfrac{7B}{9}=\dfrac{7B-7A}{9-8}=7B-7A=5\)
\(\Rightarrow\left\{{}\begin{matrix}7A=8.5=40\left(emhs\right)\\7B=9.5=45\left(emhs\right)\end{matrix}\right.\)
Câu2
Phần a
Tạm hiểu A=a {chuẩn A\(\ne a\)} vớ đề này hiểu giống nhau
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{\left(a-b\right)}{c-d}=\dfrac{\left(a+b\right)}{c+d}\)
\(\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(c-d\right)\left(c+d\right)}=\dfrac{a}{c}\dfrac{b}{d}=\dfrac{ab}{cd}\)
phầnb
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)\)\(M=\left(\dfrac{a+b}{c}\right)\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)=2.2.2=8\)
Câu 1: Thực hiện phép tính :
a) \(2.\left(\dfrac{-2}{3}\right)^2-\dfrac{7}{2}=2.\dfrac{4}{9}-\dfrac{7}{2}\)
\(=\dfrac{8}{9}-\dfrac{7}{2}\)
\(=\dfrac{16}{18}-\dfrac{63}{18}=\dfrac{-47}{18}\)
\(b,5\dfrac{4}{13}.\dfrac{-3}{4}+3\dfrac{9}{13}.\left(-0,75\right)=\dfrac{69}{13}.\dfrac{-3}{4}+\dfrac{48}{13}.\dfrac{-3}{4}\)
\(=\left(\dfrac{69}{13}+\dfrac{48}{13}\right).\dfrac{-3}{4}\)
\(=\dfrac{117}{13}.\dfrac{-3}{4}\)
\(=9.\dfrac{-3}{4}=\dfrac{-27}{4}\)
\(c,\left(-1\right)^{2017}+\left|\dfrac{-1}{13}\right|+\sqrt{\dfrac{144}{169}}=-1+\dfrac{1}{13}+\dfrac{12}{13}\)
\(=-1+\dfrac{13}{13}\)
\(=-1+1=0\)
Câu 3: Tìm x, biết:
a)\(\dfrac{3}{5}-x=25\)
\(x=\dfrac{3}{5}-\dfrac{125}{5}\)
\(x=\dfrac{-122}{5}\)
b)\(\dfrac{2}{3}\left|x-1\right|+\dfrac{1}{4}=\dfrac{5}{3}\)
\(\dfrac{2}{3}\left|x-1\right|=\dfrac{20}{12}-\dfrac{3}{12}\)
\(\dfrac{2}{3}\left|x-1\right|=\dfrac{17}{12}\)
\(\left|x-1\right|=\dfrac{17}{12}:\dfrac{2}{3}\)
\(\left|x-1\right|=\dfrac{17}{12}.\dfrac{3}{2}\)
\(\left|x-1\right|=\dfrac{17}{8}\)
Ta có 2 TH: TH1:\(x-1=\dfrac{17}{8}\) TH2:\(x-1=\dfrac{-17}{8}\) \(x=\dfrac{17}{8}+1\) \(x=\dfrac{-17}{8}+1\) \(x=\dfrac{17}{8}+\dfrac{8}{8}=\dfrac{25}{8}\) \(x=\dfrac{-17}{8}+\dfrac{8}{8}=\dfrac{-9}{8}\) Vậy x∈\(\left\{\dfrac{25}{5};\dfrac{-9}{8}\right\}\)
Câu 1 :
a, \(=\left(\dfrac{3}{4}.\dfrac{1}{5}\right).\left(26-44\right)=\dfrac{3}{20}.\left(-18\right)=\dfrac{-27}{10}\)b,
\(=\left(-8\right).\left(-0,75\right)-0,25.4-2.\dfrac{7}{6}\)
\(=\left(-6\right)-1-\dfrac{7}{3}=-7-2\dfrac{1}{3}=-9\dfrac{1}{3}\)
Câu 2 :
a, \(\rightarrow4\dfrac{1}{3}=\dfrac{6}{0,3}.\dfrac{x}{4}\)
\(\rightarrow\dfrac{13}{3}=20.\dfrac{x}{4}\)
\(\rightarrow13.4=20.x.4\rightarrow13=20.x\\ \Rightarrow x=\dfrac{13}{20}\)
b, \(\rightarrow\)TH1:
x + 1 = 4 , 5 \(\rightarrow x=4,5-1\Rightarrow x=3,5\)
\(\rightarrow\)TH2 :
x + 1 = 4 , 5 \(\rightarrow x=-4,5-1\Rightarrow x=-5,5\)
Câu 1.
a. \(\dfrac{3}{4}.26\dfrac{1}{5}-\dfrac{3}{4}.44\dfrac{1}{5}\)
\(=\dfrac{3}{4}.\left(26\dfrac{1}{5}-44\dfrac{1}{5}\right)\)
\(=\dfrac{3}{4}.(-18)\)
\(=-13\dfrac{1}{2}\)
b.\(\left(-2\right)^3.\left(-\dfrac{3}{4}\right)-0,25:\dfrac{1}{4}-2.1\dfrac{1}{6}\)
\(=\left(-8\right).\left(-\dfrac{3}{4}\right)-\dfrac{1}{4}:\dfrac{1}{4}-2.\dfrac{7}{6}\)
\(=6-1-\dfrac{7}{3}\)
\(=5-\dfrac{7}{3}\)
\(=\dfrac{8}{3}\)
Câu 2.
a. \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(4\dfrac{1}{3}:\dfrac{x}{4}=20\)
\(\dfrac{x}{4}=4\dfrac{1}{3}:20\)
\(\dfrac{x}{4}=\dfrac{13}{60}\)
\(\dfrac{15x}{60}=\dfrac{13}{60}\)
\(\Rightarrow15x=13\)
\(x=13:15\)
\(x=\dfrac{13}{15}\)
b. \(\left|x+1\right|=4,5\)
\(\Rightarrow x+1\in\left\{\pm4,5\right\}\)
* \(x+1=-4,5\)
\(x=-4,5-1\)
\(x=-5,5\)
* \(x+1=4,5\)
\(x=4,5-1\)
\(x=3,5\)
Vậy \(x\in\left\{-5,5;3,5\right\}\)
Adu! đề cc gì v?
B1: \(\dfrac{\left(1,16-x\right).5,25}{\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right).2\dfrac{2}{17}}=75\%\Rightarrow\dfrac{\left(\dfrac{29}{25}-x\right).\dfrac{21}{4}}{\left(\dfrac{95}{9}-\dfrac{29}{4}\right).\dfrac{36}{17}}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{\left(\dfrac{29}{25}-x\right).\dfrac{21}{4}}{\dfrac{119}{36}.\dfrac{36}{17}}=\dfrac{3}{4}\Rightarrow\dfrac{\left(\dfrac{29}{25}-x\right).\dfrac{21}{4}}{7}=\dfrac{3}{4}\Rightarrow\dfrac{29}{25}-x=\dfrac{3}{4}.7:\dfrac{21}{4}=1\)
\(\Rightarrow x=\dfrac{29}{25}-1=\dfrac{4}{25}\)
B2: Đề chưa rõ :V
B3: Lười giải lắm (hihi)
Gọi số học sinh lớp 7a,7b,7c lần lượt là a,b,c
Theo đề, ta có:
\(\dfrac{a}{\dfrac{2}{3}}=\dfrac{b}{\dfrac{3}{4}}=\dfrac{c}{\dfrac{4}{5}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{2}{3}}=\dfrac{b}{\dfrac{3}{4}}=\dfrac{c}{\dfrac{4}{5}}=\dfrac{a+b+c}{\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}}=\dfrac{133}{\dfrac{133}{60}}=60\)
=>a=40; b=45; c=48
1) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2010}=\dfrac{2010}{a}=\dfrac{a+b+c+2010}{b+c+2010+a}=1\)
\(\dfrac{2010}{a}=1\Rightarrow a=2010\);
\(\dfrac{c}{2010}=1\Rightarrow c=2010\);
\(\dfrac{b}{c}=1\Rightarrow\dfrac{b}{2010}=1\Rightarrow b=2010\).
Vậy (a, b, c) = (2010; 2010; 2010)
3)
a) \(A=\sqrt{x+24}+\dfrac{4}{7}\)
Có: \(\sqrt{x+24}\ge0\forall x\in R\)
\(\Rightarrow\sqrt{x+24}+\dfrac{4}{7}\ge\dfrac{4}{7}\forall x\in R\)
\(\Rightarrow A\ge\dfrac{4}{7}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+24}=0\Rightarrow x+24=0\Rightarrow x=-24\)
Vậy GTNN của \(A=\dfrac{4}{7}\Leftrightarrow x=-24\)
b) \(B=\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\)
Có: \(\sqrt{2x+\dfrac{4}{13}}\ge0\forall x\in R\)
\(\Rightarrow\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\ge-\dfrac{13}{191}\forall x\in R\)
\(\Rightarrow B\ge-\dfrac{13}{191}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{2x+\dfrac{4}{13}}=0\)
\(\Rightarrow2x+\dfrac{4}{13}=0\)
\(\Rightarrow2x=-\dfrac{4}{13}\)
\(\Rightarrow x=-\dfrac{2}{13}\)
Vậy GTNN của \(B=-\dfrac{13}{191}\Leftrightarrow x=-\dfrac{2}{13}\)
4)
a) \(A=-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\)
Có: \(\sqrt{x+\dfrac{5}{41}}\ge0\forall x\in R\)
\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}\le0\forall x\in R\)
\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\le\dfrac{7}{12}\forall x\in R\)
\(\Rightarrow A\le\dfrac{7}{12}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+\dfrac{5}{41}}=0\)
\(\Rightarrow x+\dfrac{5}{41}=0\)
\(\Rightarrow x=-\dfrac{5}{41}\)
Vậy GTLN của \(A=\dfrac{7}{12}\Leftrightarrow x=-\dfrac{5}{41}\)
b) \(B=\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\)
Có: \(\sqrt{x-\dfrac{2}{3}}\ge0\forall x\in R\)
\(\Rightarrow-\sqrt{x-\dfrac{2}{3}}\le0\forall x\in R\)
\(\Rightarrow\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\le\dfrac{-5}{13}\forall x\in R\)
\(\Rightarrow B\le\dfrac{-5}{13}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x-\dfrac{2}{3}}=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy GTLN của \(B=\dfrac{-5}{13}\Leftrightarrow x=\dfrac{2}{3}\)
Bài 3:
a, \(x:\left(\dfrac{1}{3}-\dfrac{1}{5}\right)=\dfrac{-1}{2}\)
\(x:\left(\dfrac{5-3}{15}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{2}{15}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{2}{15}\)
\(x=\dfrac{\left(-1\right).1}{1.15}=\dfrac{-1}{15}\)
b,\(\left|x+1\right|-\dfrac{4}{5}=5\dfrac{1}{5}\)
\(\left|x+1\right|-\dfrac{4}{5}=\dfrac{26}{5}\)
\(\left|x+1\right|=\dfrac{26+4}{5}=\dfrac{30}{5}=6\)
=> \(x+1=\pm6\), ta có hai trường hợp:
Trường hợp 1:
x + 1 = 6
x = 6 - 1 = 5
Trường hợp 2:
x + 1 = -6
x = (- 6) + (- 1) = -7
Vậy x ∈ {5;-7}
Gọi số học sinh lớp 7A, 7B, 7C lần lượt là: x; y; x, biết x; y; z tỉ lệ với 10; 9; 8, ta có:
\(\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{8}\) và x - y = 5
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{8}=\dfrac{x-y}{10-9}=\dfrac{5}{1}=5\)
Suy ra:
\(\dfrac{x}{10}=5\) => x = 5 . 10 = 50
\(\dfrac{y}{9}=5\) => y = 5 . 9 = 45
\(\dfrac{x}{8}=5\) => x = 5 . 8 = 40
=> x = 50, y = 45, z = 40
Vậy lớp 7A có 50 học sinh;
lớp 7B có 45 học sinh;
lớp 7C có 40 học sinh;