1) Tìm x:
a) 2005 - x + 2006 = 2007
b) 2003 < x + 200 < 2005
c) x : 5 = 1402 ( dư 3 )
d) 409 : x = 5 ( dư 4 )
e) 30 : 6 < x : 5 < 35 : 5
Bài 2: Tìm x:
a)\(\dfrac{x-1}{27}\)=\(\dfrac{-3}{1-x}\) c)\(3\times x=2\times y\) và\(x-2\times y=8\)
b)\(\dfrac{4}{5}\)-\(\left|x-\dfrac{1}{2}\right|\)=\(\dfrac{3}{4}\) d)\(\dfrac{x-1}{2005}\)=\(\dfrac{3-y}{2006}\) và x-4009=y
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
x+2/2006+x+3/2005+x+1/2004+x+5/2003=x+4/503
\(\frac{x+6}{2001}+\frac{x+5}{2002}+\frac{x+4}{2003}=\frac{x+3}{2004}+\frac{x+2}{2005}\)+\(\frac{x+1}{2006}\)
Tìm x
1)Tim x, biet: 1-4+7-10+.............-x=-75
2)Tinh 1+2-3-4+5+6-......+2002-2003-2004+2005+2006
2)
đặt a= 1+2-3-4+5+6-........+2002-2003-2004+2005+2006
Biểu thức a có (2006-1)/1+1=2006(số hạng)
Nhóm 4 số hạng vào một nhóm ta có 2006 / 4= 501 dư 2 số hạng để ra một số đầu và một số cuối
a= 1+(2-3-4+5)+(6-7-8+9)-.........+(2002-2003-2004+2005) + 2006
a=1+0+0+......+0+2006
a=1+2006
a=2007
vậy a = 2007
2. Tìm x:
a) 4/5 + x = 2/3
b) 1/2 - x = 7/12
c) 3 và 1/2 : x = -7/2
d) 3/8 - 1/6 x =5/2
e) x + 50%x = -1,5
a: x=2/3-4/5=10/15-12/15=-2/15
b: 1/2-x=7/12
=>x=1/2-7/12=-1/12
c: =>7/2:x=-7/2
=>x=-1
d: =>1/6x=3/8-5/2=3/8-20/8=-17/8
=>x=-17/8*6=-102/8=-51/4
e: =>1,5x=-1,5
=>x=-1
giải các phương trình sau:
)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2006}+\frac{x+6}{2003}\)
tính nhanh
a, 4/7 x 5/7 + 3/7 x 5/6
b, 5/9 x 1/4 + 4/9 x 3/12
c, 7/9 x 8/5 - 7/9 x 3/5
d, 2006/2005 x 3/4 - 3/4 x 1/2005
a, \(\frac{4}{7}x\frac{5}{7}+\frac{3}{7}x\frac{5}{6}\)
= ( \(\frac{4}{7}+\frac{3}{7}\)) x \(\frac{5}{7}x\frac{5}{6}\)
= 1 x \(\frac{5}{7}x\frac{5}{6}\)
= \(\frac{5}{7}x\frac{5}{6}\)
= \(\frac{25}{42}\)
Tương tự mấy câu sau cũng làm như thế này
Tìm x biết
a)\(x^4-30x^2+31x-30=0\)
b)\(\left(x^2+x\right)^2+4\times\left(x^2+x\right)=12\)
c)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
Tìm x biết:
1).40÷x dư 4; 45÷x dư 3; 50÷x dư 2.
2).x÷3 dư 1; x÷4 dư 2; x÷5 dư 3 và x<200.
3).x-1 là ước của 6.
4).10 chia hết cho (2x+1).
5).x+13 chia hết cho x+1.
6).2x+108 chia hết cho 2x+3
\(x-1\in\left\{1;6;2;3;-1;-6;-2;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;7;3;4;0;-5;-1;-2\right\}\)
\(10⋮2x+1\)
\(\Rightarrow2x+1\in\left\{1;2;5;10;-1;-2;-5;-10\right\}\)
\(\Rightarrow2x\in\left\{0;1;4;9;-2;-6;-11\right\}\)
\(\Leftrightarrow x\in\left\{0;\frac{1}{2};2;\frac{9}{2};-1;-3;-\frac{11}{2}\right\}\)
\(x+13⋮x+1\)
\(\Leftrightarrow\left(x+1\right)+12⋮x+1\)
Do \(x+1⋮x+1\) nên \(12⋮x+1\)
\(\Rightarrow x+1\in\left\{1;12;6;2;4;3;-1;-12;-6;-2;-4;-3\right\}\)
\(\Rightarrow x\in\left\{0;11;5;1;3;2;-2;-13;-7;-3;-5;-4\right\}\)