123 + [ 54 + ( - 123 ) + 46 ]

= 123 + [ - 69 + 46 ]

= 123 + ( - 23 )

= 100 

TK

Bài làm

123 + [ 54 + ( -123 ) + 46 ]

= 123 + [ 54 - 123 + 46 ]

= 123 - 23

= 100

a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=0,1.0,5=0,05\left(mol\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Theo đề:0,2......0,05

Lập tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,05}{3}\)=> Al dư, H2SO4 hết

=> \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

=> Chọn C

b) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=\dfrac{0,05}{3}=\dfrac{1}{60}\left(mol\right)\)

=> \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{1}{\dfrac{60}{0,1}}==0,17M\)

=> Chọn A

Đây là câu a nha câu b mình đang làm bạn đợi tí

Bài 2:

Xét ΔABC vuông tại C có

\(CB=BA\cdot\sin60^0=12\cdot\dfrac{\sqrt{3}}{2}=6\sqrt{3}\left(cm\right)\)

Gọi thời gian vòi I chảy riêng đến khi đầy bể là \(x\) (giờ)

Trong 1 giờ vòi I chảy được \(\dfrac{1}{x}\) bể.

Đổi: 1 giờ 20 phút = \(\dfrac{4}{3}\) giờ

Mỗi giờ hai vòi chảy được là \(\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\) bể, vậy mỗi giờ vòi II chảy được \(\dfrac{3}{4}-\dfrac{1}{x}\) (bể)

Đổi: 10 phút = \(\dfrac{1}{6}\) (giờ), 12 phút = \(\dfrac{1}{5}\) (giờ)

Ta có phương trình: \(\dfrac{1}{6}.\dfrac{1}{x}+\dfrac{1}{5}.\left(\dfrac{3}{4}-\dfrac{1}{x}\right)=\dfrac{2}{15}\)

\(\Rightarrow\dfrac{1}{6x}+\dfrac{3}{20}-\dfrac{1}{5x}=\dfrac{2}{15}\Rightarrow-\dfrac{1}{30x}=-\dfrac{1}{60}\Rightarrow x=2\)

Vậy vòi I chảy riêng trong 2 giờ sẽ đầy bể.

Mỗi giờ vòi II chảy được là \(\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\) bể, nên vòi II chảy riêng trong 4 giờ thì đầy bể.

mình cammonn ạ

x : 3 dư 2

x : 5 dư 1

→ x + 4 chia hết cho 3 và 5

→ x + 4 € BC ( 3, 5 )

Ta có: 3 . 5 = 15

→ BC ( 3, 5 ) = B ( 15 ) = {0;15;30;45;...}

Dựa vào các điều kiện trên, ta kết luận: Vậy x € { 15;30 }

a)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y+z}{3+5+6}=\dfrac{98}{14}=7\)

\(+)\)\(\dfrac{x}{3}=7\Rightarrow x=7\times3=21\)

\(+)\)\(\dfrac{y}{5}=7\Rightarrow y=7\times5=35\)

\(+)\)\(\dfrac{z}{6}=7\Rightarrow z=7\times6=42\)

Vậy \(x=21;y=35;z=42\)

b)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x+y-z}{4+5-2}=\dfrac{21}{7}=3\)

\(+)\)\(\dfrac{x}{4}=3\Rightarrow x=3\times4=12\)

\(+)\)\(\dfrac{y}{5}=3\Rightarrow y=3\times5=15\)

\(+)\)\(\dfrac{z}{2}=3\Rightarrow z=3\times2=6\)

Vậy \(x=12;y=15;z=6\)

c)

Ta có : 

\(x:y:z=5:\left(-6\right):7\) và \(x-y-z=16\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{-6}=\dfrac{z}{7}\) và \(x-y-z=16\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{5}=\dfrac{y}{-6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-\left(-6\right)-7}=\dfrac{16}{4}=4\)

\(+)\)\(\dfrac{x}{5}=4\Rightarrow x=4\times5=20\)

\(+)\)\(\dfrac{y}{-6}=4\Rightarrow y=4\times\left(-6\right)=-24\)

\(+)\)\(\dfrac{z}{7}=4\Rightarrow z=4\times7=28\)

Vậy \(x=20;y=-24;z=28\)

d)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+z}{2+4}=\dfrac{18}{6}=3\)

\(+)\)\(\dfrac{x}{2}=3\Rightarrow x=3\times2=6\)

\(+)\)\(\dfrac{y}{3}=3\Rightarrow y=3\times3=9\)

\(+)\)\(\dfrac{z}{4}=3\Rightarrow z=3\times4=12\)

Vậy \(x=6;y=9;z=12\)

e)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y}{5-6}=\dfrac{36}{-1}=-36\)

\(+)\)\(\dfrac{x}{5}=-36\Rightarrow x=-36\times5=-180\)

\(+)\)\(\dfrac{y}{6}=-36\Rightarrow y=-36\times6=-216\)

\(+)\)\(\dfrac{z}{7}=-36\Rightarrow z=-36\times7=-252\)

Vậy \(x=-180;y=-216;z=-252\)

a: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y+z}{3+5+6}=\dfrac{98}{14}=7\)

=>x=21; y=35; z=42

b: x/4=y/5=z/2 và x+y-z=21

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x+y-z}{4+5-2}=\dfrac{21}{7}=3\)

=>x=12; y=15; z=6

c: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{5}=\dfrac{y}{-6}=\dfrac{z}{7}=\dfrac{x-y-z}{5+6-7}=\dfrac{16}{4}=4\)

=>x=20; y=-24; z=28

d: Áp dụng tính chất của DTSBN, ta được:

x/2=y/3=z/4=(x+z)/(2+4)=18/6=3

=>x=6; y=9; z=12

Bài 4:

a: \(\sqrt{1,6}\cdot\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}\)

\(=\sqrt{1,6\cdot250}+\sqrt4\)

\(=\sqrt{400}+2=20+2=22\)

b: \(\sqrt{1\frac34\cdot2\frac27\cdot5\frac49}\)

\(=\sqrt{\frac74\cdot\frac{16}{7}\cdot\frac{49}{9}}=\sqrt{\frac{16}{4}\cdot\frac{49}{9}}=2\cdot\frac73=\frac{14}{3}\)

c: \(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)

\(=20\sqrt{20}-15\sqrt{45}+5\sqrt5\)

\(=40\sqrt5-45\sqrt5+5\sqrt5=0\)

d: \(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)

\(=\sqrt{25}-\sqrt9+2\cdot\sqrt{16}\)

\(=5-3+2\cdot4\)

=2+8

=10

e: \(\frac{2\sqrt8-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt5+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\frac{4\sqrt2-2\sqrt3}{\sqrt6\left(\sqrt3-2\sqrt2\right)}-\frac{\sqrt5+\sqrt{27}}{\sqrt6\left(\sqrt5+\sqrt{27}\right)}\)

\(=\frac{2\left(2\sqrt2-\sqrt3\right)}{-\sqrt6\left(2\sqrt2-\sqrt3\right)}-\frac{1}{\sqrt6}=-\frac{2}{\sqrt6}-\frac{1}{\sqrt6}=-\frac{3}{\sqrt6}=\frac{-3\sqrt6}{6}=-\frac{\sqrt6}{2}\)

f: \(\frac{3+2\sqrt3}{\sqrt3}+\frac{2+\sqrt2}{\sqrt2+1}-\left(\sqrt2+\sqrt3\right)\)

\(=2+\sqrt3+\frac{\sqrt2\left(\sqrt2+1\right)}{\sqrt2+1}-\sqrt2-\sqrt3\)

\(=2-\sqrt2+\sqrt2\)

=2

bài 1:

\(a.\sqrt{25 . 144}=\sqrt{25}.\sqrt{144}=5.12=60\)

\(b.\sqrt{45 . 80}=\sqrt{9 . 5 . 5 . 16}=\sqrt{9 . 25 . 16}=\sqrt{9}.\sqrt{25}.\sqrt{16}=3.5.4=60\)

\(c.\sqrt{52}.\sqrt{13}=\sqrt{52 . 13}=\sqrt{4 . 13 . 13}=\sqrt{4 . 13^2}=\sqrt{4}.\sqrt{13^2}=2.13=26\)

\(d.\sqrt{7}.\sqrt{28}=\sqrt{7 . 28}=\sqrt{7 . 7 . 4}=\sqrt{7^2 . 4}=\sqrt{7^2}.\sqrt{4}=7.2=14\)

\(e.\sqrt{1 \frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{\sqrt{25}}{\sqrt{16}}=\frac{5}{4}\)

\(f.\sqrt{\frac{25}{64}}=\frac{\sqrt{25}}{\sqrt{64}}=\frac{5}{8}\)

\(g.\frac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\frac{12,5}{0,5}}=\sqrt{25}=5\)

\(h.\frac{\sqrt{230}}{\sqrt{2,3}}=\sqrt{\frac{230}{2,3}}=\sqrt{100}=10\)

bài 2:

\(a.\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{50}{3}}-\sqrt{24}\right).\sqrt{6}\)

\(= \sqrt{\frac{2}{3}} . \sqrt{6} + \sqrt{\frac{50}{3}} . \sqrt{6} - \sqrt{24} . \sqrt{6}\)

\(= \sqrt{\frac{2}{3} . 6} + \sqrt{\frac{50}{3} . 6} - \sqrt{24 . 6}\)

\(= \sqrt{4} + \sqrt{100} - \sqrt{144}\)

\(=2+10-12=0\)

b. \(\sqrt{3 + \sqrt{5}}.\sqrt{2}=\sqrt{(3 + \sqrt{5}) . 2}\)

\(=\sqrt{6 + 2\sqrt{5}}=\sqrt{5 + 2\sqrt{5} + 1}\)

\(=\sqrt{(\sqrt{5} + 1)^2}=\vert{}\sqrt{5}+1\vert{}=\sqrt{5}+1\)

\(c.\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\sqrt{\frac{4}{3}}\right).\sqrt{12}\)

\(= \sqrt{\frac{3}{4}} . \sqrt{12} - \sqrt{3} . \sqrt{12} + 5\sqrt{\frac{4}{3}} . \sqrt{12}\)

\(= \sqrt{\frac{3}{4} . 12} - \sqrt{3 . 12} + 5\sqrt{\frac{4}{3} . 12}\)

\(= \sqrt{9} - \sqrt{36} + 5\sqrt{16}\)

\(= 3 - 6 + 5 . 4\)

\(=3-6+20=17\)

\(d.\sqrt{3 - \sqrt{5}}.\sqrt{8}=\sqrt{3 - \sqrt{5}}.\sqrt{2}.\sqrt{4}\)

\(=\sqrt{(3 - \sqrt{5}) . 2}.2=2\sqrt{6 - 2\sqrt{5}}\)

\(=2\sqrt{5 - 2\sqrt{5} + 1}=2\sqrt{(\sqrt{5} - 1)^2}\)

\(=2\vert{}\sqrt{5}-1\vert{}=2(\sqrt{5}-1)=2\sqrt{5}-2\)

bài 3:

\(a.\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)

\(= \sqrt{\frac{1}{7}} : \sqrt{7} - \sqrt{\frac{16}{7}} : \sqrt{7} + \sqrt{7} : \sqrt{7}\)

\(= \sqrt{\frac{1}{7} : 7} - \sqrt{\frac{16}{7} : 7} + 1\)

\(= \sqrt{\frac{1}{49}} - \sqrt{\frac{16}{49}} + 1\)

\(=\frac{1}{7}-\frac{4}{7}+1=-\frac{3}{7}+1=\frac{4}{7}\)

\(b.\sqrt{36 - 12\sqrt{5}}:\sqrt{6}=\sqrt{\frac{36 - 12\sqrt{5}}{6}}\)

\(=\sqrt{6 - 2\sqrt{5}}=\sqrt{5 - 2\sqrt{5} + 1}\)

\(=\sqrt{(\sqrt{5} - 1)^2}=\vert{}\sqrt{5}-1\vert{}=\sqrt{5}-1\)

\(c.\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(= \sqrt{\frac{1}{3}} : \sqrt{3} - \sqrt{\frac{4}{3}} : \sqrt{3} + \sqrt{3} : \sqrt{3}\)

\(= \sqrt{\frac{1}{3} : 3} - \sqrt{\frac{4}{3} : 3} + 1\)

\(=\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+1=\frac{1}{3}-\frac{2}{3}+1\)

\(=-\frac{1}{3}+1=\frac{2}{3}\)

\(e.\sqrt{3 - \sqrt{5}}:\sqrt{2}=\sqrt{\frac{3 - \sqrt{5}}{2}}\)

\(=\sqrt{\frac{6 - 2\sqrt{5}}{4}}=\frac{\sqrt{6 - 2\sqrt{5}}}{\sqrt{4}}\)

\(=\frac{\sqrt{5 - 2\sqrt{5} + 1}}{2}=\frac{\sqrt{(\sqrt{5} - 1)^2}}{2}\)

\(=\frac{\vert{}\sqrt{5} - 1\vert{}}{2}=\frac{\sqrt{5} - 1}{2}\)

bài 4:

\(a.\sqrt{1,6}.\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}=\sqrt{1,6 . 250}+\sqrt{\frac{19,6}{4,9}}\)

\(=\sqrt{400}+\sqrt{4}=20+2=22\)

\(b.\sqrt{1 \frac{3}{4}}.\sqrt{2 \frac{2}{7}}.\sqrt{5 \frac{4}{9}}=\sqrt{\frac{7}{4}}.\sqrt{\frac{16}{7}}.\sqrt{\frac{49}{9}}\)

\(=\sqrt{\frac{7}{4} . \frac{16}{7} . \frac{49}{9}}=\sqrt{\frac{16 . 49}{4 . 9}}=\sqrt{\frac{4 . 49}{9}}\)

\(=\frac{\sqrt{4} . \sqrt{49}}{\sqrt{9}}=\frac{2 . 7}{3}=\frac{14}{3}\)

\(c.\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)

\(= 20\sqrt{300} : \sqrt{15} - 15\sqrt{675} : \sqrt{15} + 5\sqrt{75} : \sqrt{15}\)

\(= 20\sqrt{\frac{300}{15}} - 15\sqrt{\frac{675}{15}} + 5\sqrt{\frac{75}{15}}\)

\(= 20\sqrt{20} - 15\sqrt{45} + 5\sqrt{5}\)

\(= 20\sqrt{4 . 5} - 15\sqrt{9 . 5} + 5\sqrt{5}\)

\(= 20 . 2\sqrt{5} - 15 . 3\sqrt{5} + 5\sqrt{5}\)

\(= 40\sqrt{5} - 45\sqrt{5} + 5\sqrt{5}\)

\(= (40 - 45 + 5)\sqrt{5}\)

\(=0\sqrt{5}=0\)

d. \(\left( \sqrt{325} - \sqrt{117} + 2\sqrt{208} \right) : \sqrt{13}\)

\(= \sqrt{325} : \sqrt{13} - \sqrt{117} : \sqrt{13} + 2\sqrt{208} : \sqrt{13}\)

\(= \sqrt{\frac{325}{13}} - \sqrt{\frac{117}{13}} + 2\sqrt{\frac{208}{13}}\)

\(= \sqrt{25} - \sqrt{9} + 2\sqrt{16}\)

\(=5-3+2.4=10\)

\(e.\frac{2\sqrt{8} - \sqrt{12}}{\sqrt{18} - \sqrt{48}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{30} + \sqrt{162}}=\frac{2\sqrt{4 . 2} - \sqrt{4 . 3}}{\sqrt{9 . 2} - \sqrt{16 . 3}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{6 . 5} + \sqrt{81 . 2}}\)

\(=\frac{2 . 2\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 4\sqrt{3}}.\frac{\sqrt{5} + 3\sqrt{3}}{\sqrt{6}.\sqrt{5} + 9\sqrt{2}}=\frac{4\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 4\sqrt{3}}.\frac{\sqrt{5} + 3\sqrt{3}}{\sqrt{6}(\sqrt{5} + 3\sqrt{3})}\)

\(=\frac{2(2\sqrt{2} - \sqrt{3})}{3\sqrt{2} - 4\sqrt{3}}.\frac{1}{\sqrt{6}}=\frac{2(2\sqrt{2} - \sqrt{3})}{\sqrt{6}(3\sqrt{2} - 4\sqrt{3})}\)