41 x 3 = ?
5/6 + 1/2 : 3/4
5/6 - 1/2 x 3/4
1/2 x 1/3 : 1/4
1/2 : 1/3 x 1/4
1/2 : 1/3 : 1/4
5/6+ 1/2 x 4/3
= 5/6+ 2/3
= 5/6+4/6
= 11/6
5/6- 3/8
= 20/24- 9/24
= 11/24
a) x – 36 : 18 = 12 – 15
b) 92 – (17 + x) = 72
c) 720 : [41 – (2x + 5)] = 40
d) (x + 2)3 - 23 = 41
\(a,x-36:18=12-15\\ \Rightarrow x-2=-3\\ \Rightarrow x=-1\\ b,92-\left(17+x\right)=72\\ \Rightarrow17+x=20\\ \Rightarrow x=3\\ c,720:\left[41-\left(2x+5\right)\right]=40\\ \Rightarrow41-\left(2x+5\right)=18\\ \Rightarrow2x+5=23\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ d,\left(x+2\right)^3-23=41\\ \Rightarrow\left(x+2\right)^3=64\\ \Rightarrow\left(x+2\right)^3=4^3\\ \Rightarrow x+2=4\\ \Rightarrow x=2\)
b: =>x+17=20
hay x=3
d: =>x+2=4
hay x=2
tìm x thoả mãn: \(x^3+\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}x+5=0\)
Xét \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Phương trình trên tương đương:
x3+4x+5=0
<=>x(x2-1)+5(x+1)=0
<=>x(x-1)(x+1)+5(x+1)=0
<=>(x+1)(x2-x+5)=0
<=>x+1=0 hoặc x2-x+5=0(vô nghiệm)
<=>x=-1
Vậy pt trên có nghiệm là x=-1
Bài này đi thi vio mk cũng gặp ..
bằng 1 ak
thực hiện phép tính :
a, -1/2 + 19/23+ -9/2+ 4/23 +3 /-6
b. 5/13+ -5/7 +-20/41 + 8/13 +-21/41
c.-5/7 x 19/41 + 5/7 x -21/41 + 5/7
d. 7/3 +1/2 - -3/70
Đây là phân số
Gửi đến bạn Vũ:
\(x-\frac{6}{7}=5+\frac{2}{3}.x\)
\(\Rightarrow x=5+\frac{2}{3}.x+\frac{6}{7}\)
\(\Rightarrow x=\frac{41}{7}+\frac{2}{3}x\)
\(\Rightarrow0=\frac{41}{7}+\frac{2}{3}x-x\)
\(\Rightarrow0=\frac{41}{7}+x\left(-\frac{1}{3}\right)\)
\(\Rightarrow-\frac{41}{7}=x.\left(-\frac{1}{3}\right)\)
\(\Rightarrow x=\frac{123}{7}\)
gửi đến bạn vũ thì gửi cho bạn đấy chứ đăng lên đây làm gì.
Đỗ Công Tùng em đăng lên đây k để câu sub hay câu tk gì hết bởi vì em đăng bài chớ đâu đăng câu trả lời :D xàm quá phắn giùm em đê
tìm x thuộc z: 27/23+-4/23+1/2+-4/8<x<7/3+13/41+28/41
\(\frac{27}{23}+\frac{-4}{23}+\frac{1}{2}+\frac{-4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)
\(\Rightarrow\frac{27}{23}-\frac{4}{23}+\frac{1}{2}-\frac{4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)
\(\Rightarrow1+0< x< \frac{7}{3}+\frac{3}{3}\)
\(\Rightarrow1< x< \frac{10}{3}\)
\(\Rightarrow1< x< 3,333333333\)
\(\Rightarrow x\in\left\{2;3\right\}\)
Vậy : ....
ta co : \(\frac{27}{23}+\frac{-4}{23}+\frac{1}{2}+\frac{-4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)
=> \(1< x< \frac{10}{3}\)
vi x la so nguyen => \(1< x\le3\)
con lai ban tu lam
tìm tập hợp các số nguyên x mà x thuộc z biết 15/41 + -138/41<x< 1/2+1/3+1/6
Tính nhanh: 37. (41 + 63) – 63. (37 – 41)
Tìm x, biết: 5. (4 - x) = 15 + 2. (3x - 3)
Ta có:
37.41+37.63-63.37+63.41
=41(37+63)
=41.100
=4100
Ta có:
20-5x=15+6x-6
<=>20-5x-15-6x+6=0
<=>11-11x=0
<=>11x=11
<=>x=1
Vậy x=1
x+1/x-3-41/x+3+x2-22/9-x2=0
\(\dfrac{x+1}{x-3}-\dfrac{41}{x+3}+\dfrac{x^2+22}{9-x^2}=0\left(ĐKXĐ:x\ne3;x\ne-3\right)\\ \Leftrightarrow\dfrac{x+1}{x-3}-\dfrac{41}{x+3}-\dfrac{x^2+22}{x^2-9}=0\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-41\left(x-3\right)-x^2-22}{x^2-9}=0\\ \Leftrightarrow x^2+4x+3-41x+123-x^2-22=0\\ \Leftrightarrow-37x+104=0\\ \Leftrightarrow-37x=-104\\ \Leftrightarrow x=\dfrac{104}{37}\left(tmđk\right)\)
Vậy \(x=\dfrac{104}{37}\) là nghiệm của pt.
Tính nhanh :
70/3 x ( 39/30 + 39/42 ) - 246/7 : ( 41/56 + 41/72 )
57/20 - 26/15 + 139/20 : 3
39/4 + 2/3 x ( 11 - 23/4 )
( 1 - 1/2 ) x ( 1 - 1/3) x ( 1 - 1/4 ) x ....( 1 - 1/2004 )
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)