Mỗi thùng có số lít dầu là :: ( 140 - 20 ) : 8 = 15 ( lít )

Nếu lấy 5 thùng và bớt 15 lít dầu thì có số lít dầu là : 

15 x 5 - 15 = 60 ( lít )

                  Đáp số : 60 lít dầu

các bạn giúp mik với chiều nay đi học r mà chưa làm bài :<

Bài 1: 

\(10\sqrt{\frac{1}{5}}-\frac{5}{\sqrt{5}}+\frac{1}{2-\sqrt{5}}\)

\(=10.\sqrt{\frac{5}{5^2}}-\sqrt{5}+\frac{2+\sqrt{5}}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)

\(=10.\frac{1}{5}.\sqrt{5}-\sqrt{5}+\frac{2+\sqrt{5}}{4-5}\)

\(=2\sqrt{5}-\sqrt{5}-\left(2+\sqrt{5}\right)\)

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

Bài 2: 

\(ĐKXĐ:x\ge-1\)

\(\sqrt{9x+9}-2\sqrt{25x+25}=-14\)

\(\Leftrightarrow\sqrt{9\left(x+1\right)}-2\sqrt{25\left(x+1\right)}=-14\)

\(\Leftrightarrow3\sqrt{x+1}-2.5\sqrt{x+1}=-14\)

\(\Leftrightarrow3\sqrt{x+1}-10\sqrt{x+1}=-14\)

\(\Leftrightarrow-7\sqrt{x+1}=-14\)

\(\Leftrightarrow\sqrt{x+1}=2\)

\(\Leftrightarrow x+1=4\)\(\Leftrightarrow x=3\)( thỏa mãn ĐKXĐ )

Vậy \(x=3\)

Ta có: \(a^2+b=b^2+c\Rightarrow a^2-b^2=c-b\Rightarrow\left(a+b\right)\left(a-b\right)-\left(a-b\right)=c-a\Rightarrow\left(a+b-1\right)\left(a-b\right)=c-a\)(1)\(b^2+c=c^2+a\Rightarrow b^2-c^2=a-c\Rightarrow\left(b+c\right)\left(b-c\right)-\left(b-c\right)=a-b\Rightarrow\left(b+c-1\right)\left(b-c\right)=a-b\)(2)\(c^2+a=a^2+b\Rightarrow c^2-a^2=b-a\Rightarrow\left(c+a\right)\left(c-a\right)-\left(c-a\right)=b-c\Rightarrow\left(c+a-1\right)\left(c-a\right)=b-c\)(3)

Nhân ba vế của ba đẳng thức (1), (2), (3), ta được:\(\left(a+b-1\right)\left(a-b\right)\left(b+c-1\right)\left(b-c\right)\left(c+a-1\right)\left(c-a\right)=\left(c-a\right)\left(a-b\right)\left(b-c\right)\Rightarrow\left(a+b-1\right)\left(b+c-1\right)\left(c+a-1\right)=1\)(Do a, b, c đôi mội khác nhau nên \(a-b,b-c,c-a\ne0\) )

Ta có : 

a² + b = b² + c <=> a² - b² = c - b <=> ( a + b ) ( a - b ) = c - b

<=> \(a+b=\frac{c-b}{a-b}\)<=> \(a+b-1=\frac{c-a}{a-b}\)

b² + c = c² + a <=> b² - c² = a - c <=> ( b + c ) ( b - c ) = a - c

<=> \(b+c=\frac{a-c}{b-c}\)<=> \(b+c-1=\frac{a-b}{b-c}\)

a² + b = c² + a <=> a² - c² = a - b <=> ( a + c ) ( a - c ) = a - b

<=> \(a+c=\frac{a-b}{a-c}\)<=> \(a+c-1=\frac{c-b}{a-c}\)

Suy ra :

( a + b - 1 ) ( a + c - 1 ) ( a + c - 1 ) = \(\frac{c-a}{a-b}.\frac{a-b}{b-c}.\frac{c-b}{a-c}=-\frac{a-c}{a-b}.\frac{a-b}{b-c}.\left(-\frac{b-c}{a-c}\right)=1\)

Mình làm theo đề bài là rút gọn nha!

\(\left(x+2\right)\cdot\left(x^2-2x+4\right)-2\cdot\left(x+1\right)\cdot\left(1-x\right)\)

\(=x^3+8-2\cdot\left(1-x^2\right)\)

\(=x^3+8-2+2x^2\)

\(=x^3+2x^2+6\)

=5 * 32 + 35 *4 +41 * 65 + 23* 65

=23 * (5+65) +41 * (65+35)

=23* 70 +41 *100

=1610+4100

=5710

Bài 1 :

a) 1800,5 gấp 18,005 số lần là : 1800,5 / 18,005 = 100 ( lần )

b) 5486,7 -> 54,867; 6789,1 -> 67,891; 2015 -> 20,15; 63845,6 -> 638,456

a, \(\left(\frac{1}{2}-\frac{1}{3}\right)^2-\frac{1}{4}\)

\(=\left(\frac{1}{6}\right)^2-\frac{1}{4}=\frac{1}{36}-\frac{1}{4}=\frac{1}{36}-\frac{9}{36}\)

\(=-\frac{8}{36}=-\frac{2}{9}\)

b. \(\frac{1}{2}\cdot\sqrt{64}-\sqrt{\frac{25}{49}}+1^{2020}\)

\(=\frac{1}{2}\cdot8-\frac{5}{7}-1=4-\frac{5}{7}-1\)

\(=3-\frac{5}{7}=\frac{21}{7}-\frac{5}{7}=\frac{16}{7}\)

Trước hết, ta xét bất đẳng thức phụ sau: \(m^2-mn+n^2\ge\frac{1}{3}\left(m^2+mn+n^2\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}m^2-\frac{4}{3}mn+\frac{2}{3}n^2\Leftrightarrow\frac{2}{3}\left(m-n\right)^2\ge0\)*đúng*

Xét biểu thức\(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)

Tương tự: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)

\(\Rightarrow\left(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^3}\right)+\left(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}\right)+\left(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}\right)=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)\(\Rightarrow2VT=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+\frac{\left(b+c\right)\left(b^2-bc+c^2\right)}{b^2+bc+c^2}+\frac{\left(c+a\right)\left(c^2-ca+a^2\right)}{c^2+ca+a^2}\)\(\ge\frac{\left(a+b\right)\frac{1}{3}\left(a^2+ab+b^2\right)}{a^2+ab+b^2}+\frac{\left(b+c\right)\frac{1}{3}\left(b^2+bc+c^2\right)}{b^2+bc+c^2}+\frac{\left(c+a\right)\frac{1}{3}\left(c^2+ca+a^2\right)}{c^2+ca+a^2}=\frac{2\left(a+b+c\right)}{3}\Rightarrow VT\ge\frac{a+b+c}{3}\)Đẳng thức xảy ra khi a = b = c