a) (x²+1)²-(2x+100)²=0
b) (x/2+1)³-x³/2-4=0
c) 3-x/2007-1=2-x/2008-x/2009
(x^2+1)^2-(2x+100)^2=0
(x/2 +1)^3-x^3/2 -4=0
3-x/2007 -1=2-x/2008-x/2009
\(\left(x^2+1\right)^2-\left(2x+100\right)^2=0\)
\(\Leftrightarrow\left(x^2+1-2x-100\right)\left(x^2+1-2x+100\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-99=0\\x^2-2x+101=0\left(loại\right)\end{cases}}\)
\(\Leftrightarrow\left(x-1\right)^2=100\)
\(\Leftrightarrow\orbr{\begin{cases}x=11\\x=-9\end{cases}}\) ( thỏa mãn )
a ) \(4\left(x+5\right)-3\left|2x-1\right|=0\)
b ) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
c ) \(x^4+4x^2-5=0\)
a ) \(4\left(x+5\right)-3\left|2x-1\right|=0\)
\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)
\(\Leftrightarrow\left|2x-1\right|=\frac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=\frac{4}{3}\left(x+5\right)\\2x-1=-\frac{4}{3}\left(x+5\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=\frac{4}{3}x+\frac{20}{3}\\2x-1=-\frac{4}{3}x-\frac{20}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=-\frac{23}{3}\\\frac{2}{3}x=-\frac{17}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{23}{2}\left(l\right)\\x=-\frac{17}{10}\left(n\right)\end{cases}}\)
Vậy \(x=-\frac{17}{10}\)
b ) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
\(\Leftrightarrow\frac{2-x}{2007}+1=\left(\frac{1-x}{2008}+1\right)+\left(1-\frac{x}{2009}\right)\)
\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}=\frac{2009-x}{2009}\)
\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\right)\)
\(\Leftrightarrow x=2019\)
Vậy phương trình có nghiệm \(x=2019\)
c ) \(x^4+4x^2-5=0\)
\(\Leftrightarrow x^4-x^2+5x^2-5=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\left(l\right)\\x=1\end{cases}}\)
\(x=-1\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt !!!
4(x+5)-3|2x-1|=0
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
x4+4x2-5=0
\(x^4+4x^2-5=0\)
\(\Leftrightarrow x^4-x^2+5x^2-5=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\left(l\right)\\x=1\\x=-1\end{matrix}\right.\)
\(4\left(x+5\right)-3\left|2x-1\right|=0\)
\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}\left(x+5\right)\\2x-1=-\dfrac{4}{3}\left(x+5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}x+\dfrac{20}{3}\\2x-1=-\dfrac{4}{3}x-\dfrac{20}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{23}{3}\\\dfrac{2}{3}x=-\dfrac{17}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(l\right)\\x=-\dfrac{17}{10}\left(n\right)\end{matrix}\right.\)
Vậy: \(x=-\dfrac{17}{10}\)
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2-x}{2007}+1=\left(\dfrac{1-x}{2008}+1\right)+\left(1-\dfrac{x}{2009}\right)\\ \Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\\ \Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\\ \Leftrightarrow2009-x=0\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\ne0\right)\\ \Leftrightarrow x=2019\)
Vậy phương trình có nghiệm \(x=2019\)
giải phương trình:
a) \(\left(\dfrac{x}{2}+1\right)^3-\dfrac{x^3}{2}-4=0\)
b) \(\dfrac{3-x}{2007}+1=\dfrac{2-x}{2008}-\dfrac{x}{2009}\)
Lời giải:
a)
PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)
\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)
\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)
\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)
\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)
b) Bạn kiểm tra lại xem có sai đề không?
1.Tìm x biết:
a) 7x+2+7x+1+7/57=52x+52x+1+52x+3/131.
b) (4x-3)2=(4x-3)3.
c) |5(2x+3)|+|2(2x+3)|+|2x+3|=16.
d) |x2+|6x-2||=x2+4.
2.Tính tỉ số A/B,biết:
A=1/2+1/3+1/4+...+1/2007+1/2008+1/2009.
B=2008/1+2007/2+2006/3+...+2/2007+2/2008.
a,| x+13/7 | + | y+2009/2008 | + | z-2007 |=0
b,| x-1/2 | + | y+3/2 | + | z-5/2 | \(\le\)0
c,| x+11/17 | + | x+2/17 | + | x+4/17 | =42
1. tích các nghiệm cua phương trình(x-2009)(x-2008)(x-2007)...(x-1)x(x+1)...(x+2010)=0
2.cho abc=2008 . khi do gia tri cua bieu thuc
\(\frac{2008a}{\left(ab+2008a+2008\right)}+\frac{b}{bc+b+2008}+\frac{c}{ac+c+1}=0\)
3.gia tri cua bieu thuc Q=(3+1)(32+1)(34+1)...(33994+1)
3) Q=(3+1)(3^2+1)(3^4+1)....(3^3994+1)
=(3-1)(3+1)(3^2+1)(3^4+1)...(3^3994+1)
=(3^2-1)(3^2+1)(3^4+1)...(3^3994+1)
=(3^4-1)(3^4+1)...(3^3994+1)
=.........
=(3^3994-1)(3^3994+1)
=3^7988-1
a,\(\left(x-1\right)\cdot\left(x+2\right)>0\)
b,\(\left(\frac{1}{2}-x\right)\cdot\left(2x-1\right)< 0\)
c,\(\frac{x-1}{2007}+\frac{x-2}{2008}=\frac{x-3}{2009}+\frac{x-4}{2010}\)
d,\(\left(2x-1\right)\cdot\left(2-x\right)>0\)
Tìm x
bạn nào bít thì giúp mình nha!!!
a, \(\left(x-1\right).\left(x+2\right)\)\(>0\Rightarrow\orbr{\begin{cases}x-1< 0;x+2< 0\left(loai\right)\Rightarrow x< 1\\x-1>0;x+2>0\Rightarrow x>1;x>-2\end{cases}}\)
=> -2 < x < 1
Câu b và câu d làm tương tự nha bạn(Câu b thì xét khác dấu)
a) a= 2 và 1
b) = 7
c= 5600 và 7899
d 5 và 6
Bài 1) giải phương trình sau a) 4x-20=0
b)3-2x = 3(x+1)-x-2
C) x+2/2008 + x+3/2007 + x+4/2006 + x+2028/6 = 0
d) 2x(x+3)+5(x+3)= 0
bài2 tìm điều kiện xác định của phương trình sau: 2 /x - 1 = 1 /x + l
bài6 giải phương trình
1/x-1 + 1/x-2 = 1/ x+2 + 1/x+1.
Bài 1:
a) Ta có: 4x-20=0
\(\Leftrightarrow4\left(x-5\right)=0\)
mà \(4\ne0\)
nên x-5=0
hay x=5
Vậy: x=5
b) Ta có: 3-2x=3(x+1)-x-2
\(\Leftrightarrow3-2x=3x+3-x-2\)
\(\Leftrightarrow3-2x=2x+1\)
\(\Leftrightarrow3-2x-2x-1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow2\left(1-2x\right)=0\)
mà \(2\ne0\)
nên 1-2x=0
\(\Leftrightarrow2x=1\)
hay \(x=\frac{1}{2}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)
c) Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
mà \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
nên x+2010=0
hay x=-2010
Vậy: Tập nghiệm S={-2010}
d) Ta có: 2x(x+3)+5(x+3)=0
\(\Leftrightarrow\left(x+3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-3;\frac{-5}{2}\right\}\)
Bài 2:
ĐKXĐ: \(x\notin\left\{-1;1\right\}\)
Bài 6:
ĐKXĐ: \(x\notin\left\{1;2;-1;-2\right\}\)
Ta có: \(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\Leftrightarrow\frac{x-2}{\left(x-1\right)\left(x-2\right)}+\frac{x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1}{\left(x+1\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{x-2+x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1+x+2}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{2x-3}{\left(x-1\right)\left(x-2\right)}-\frac{2x+3}{\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{\left(2x-3\right)\left(x^2+3x+2\right)}{\left(x^2-1\right)\left(x^2-4\right)}-\frac{\left(2x+3\right)\left(x^2-3x+2\right)}{\left(x^2-1\right)\left(x^2-4\right)}=0\)
\(\Leftrightarrow2x^3+3x^2-5x-6-\left(2x^3-3x^2-5x+6\right)=0\)
\(\Leftrightarrow2x^3+3x^2-5x-6-2x^3+3x^2+5x-6=0\)
\(\Leftrightarrow6x^2-12=0\)
\(\Leftrightarrow6x^2=12\)
\(\Leftrightarrow x^2=2\)
hay \(x=\pm\sqrt{2}\)
Vậy: Tập nghiệm \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)