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Tính:
\(3.\left(x-2\right)-4.\left(2x+1\right)-5.\left(2x+3\right)=50\)
\(\Rightarrow3x-6-\left(8x+4\right)-\left(10x+15\right)=50\)
\(\Rightarrow3x-6-8x-4-10x-15=50\)
\(\Rightarrow-15x-25=50\)
\(\Rightarrow-15x=50+25\)
\(\Rightarrow-15x=75\)
\(\Rightarrow x=75:\left(-15\right)\)
\(\Rightarrow x=-5.\)
Vậy \(x=-5.\)
Chúc bạn học tốt!
\(3x^2-3xy-y-5x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)-y-5x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)+x-y-6x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)+\left(x-y\right)-6x=-20\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-6x=-20\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-6x-2=-22\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-\left(6x+2\right)=-22\)
\(\Rightarrow\left(x-y\right)\left(3x+1\right)-2\left(3x+1\right)=-22\)
\(\Rightarrow\left(3x+1\right)\left(x-y-2\right)=-22\)
Ta có bảng sau:
| \(3x+1\) | \(-1\) | \(1\) | \(-22\) | \(22\) |
| \(x\) | \(x\notin Z\) | \(0\) | \(x\notin Z\) | \(7\) |
| \(x-y-2\) | \(-22\) | \(-1\) | ||
| \(y\) | \(-20\) | \(6\) |
Vậy ta có 2 bộ (x,y) là (0;-20) và (7;6)
Chúc bạn học tốt!
Ta có : \(\frac{x+1}{5}=\frac{x-2}{4}\)
=> 4(x + 1) = 5(x - 2)
=> 4x + 4 = 5x - 10
=> 4x - 5x = -10 - 4
=> -x = -14
=> x = 14
Thay x = 14 vào ta có : \(\frac{14-2}{4}=\frac{5y+1}{28}\Rightarrow\frac{84}{28}=\frac{5y+1}{28}\)
=> 5y + 1 = 84
=> 5y = 83
=> y = 83/5
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
b: \(ab\cdot bc\cdot ac=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow\left(abc\right)^2=\dfrac{1}{4}\)
Trường hợp 1: abc=1/2
\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{1}{2}:\dfrac{1}{2}=1\\a=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{3}{4}\\b=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)
Trường hợp 2: abc=-1/2
\(\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a=-\dfrac{3}{4}\\b=-\dfrac{2}{3}\end{matrix}\right.\)
c: Theo đề, ta có: \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{1}\\\dfrac{y-2}{3}=\dfrac{z-3}{4}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot6-3\cdot6+3\cdot4}=\dfrac{45}{6}=\dfrac{15}{2}\)
Do đó: x-1=45; y-2=45/2; z-3=30
=>x=46; y=49/2; z=33
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!