\(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
giúp em mai em thi rồi
Hãy giải phương trình sau!
a,\(\frac{x+\frac{x+1}{5}}{3}=1-\frac{2x-\frac{1-2x}{34}}{5}\)
b,\(\frac{^{\left(2x+1\right)^2}}{5}-\frac{^{\left(x-1\right)^2}}{3}=\frac{7x^2-14x-5}{15}\)
c,\(\frac{2x-1}{2}+\frac{2x+1}{3}=\frac{2x+7}{6}+\frac{2x+9}{7}\) ( gợi ý : thêm hoặc bớt 2 )
a) \(\frac{x+\frac{x+1}{5}}{3}=1-\frac{2x-\frac{1-2x}{34}}{5}\)
\(\Leftrightarrow\frac{\frac{5x+x+1}{5}}{3}=1-\frac{\frac{68x-1+2x}{34}}{5}\)
\(\Leftrightarrow\frac{6x+1}{15}=1-\frac{70-1}{170}\)
\(\Leftrightarrow\frac{6x+1}{15}+\frac{70x-1}{170}-1=0\)
\(\Leftrightarrow\frac{34\left(6x+1\right)+3\left(70x-1\right)-510}{510}=0\)
\(\Leftrightarrow204x+34+210x-3-510=0\)
\(\Leftrightarrow414x-479=0\)
\(\Leftrightarrow x=\frac{479}{414}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{479}{414}\right\}\)
bạn ơi bạn làm được câu c chưa
b) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{4x^2+4x+1}{5}-\frac{x^2-2x+1}{3}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5}{15}=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow x=-\frac{1}{12}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{12}\right\}\)
c) \(\frac{2x-1}{2}+\frac{2x+1}{3}=\frac{2x+7}{6}+\frac{2x+9}{7}\)
\(\Leftrightarrow\left(\frac{2x-1}{2}-2\right)+\left(\frac{2x+1}{3}-2\right)=\left(\frac{2x+7}{6}-2\right)+\left(\frac{2x+9}{7}-2\right)\)
\(\Leftrightarrow\frac{2x-5}{2}+\frac{2x-5}{3}=\frac{2x-5}{6}+\frac{2x-5}{7}\)
\(\Leftrightarrow\frac{2x-5}{2}+\frac{2x-5}{3}-\frac{2x-5}{6}-\frac{2x-5}{7}=0\)
\(\Leftrightarrow\left(2x-5\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}-\frac{1}{7}\right)=0\)
Mà \(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}-\frac{1}{7}\ne0\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
Bài 1 giải các phương trình sau
a, \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
b, \(\left(x+2\right)^2+\left(x-1\right)\left(x+3\right)=2\left(x-4\right)\left(x+4\right)\)
c, \(\frac{3}{x-1}=\frac{3x+2}{1-x^2}-\frac{4}{x+1}\)
d, \(\frac{1}{x+1}+\frac{2x^2+1}{x^3+1}+\frac{2x^3-2x^2}{x^2-x+1}=2x\)
Bài 2 : Giải phương trình \(x^4+3x^3+6x+4=0\)
các bạn ơi ! giúp mik với đi ! mai kt rồi
a,<=>\(\frac{20\left(1-2x\right)+6x}{12}\)=\(\frac{9\left(x-5\right)-24}{12}\)
=> 20-40x+6x = 9x-45-24
<=> -40x+6x-9x = -20-45-24
<=> -43x = -89
<=> x = \(\frac{89}{43}\)
c,ĐKXĐ :x\(\ne\pm1\)
<=>\(\frac{3\left(x+1\right)}{x^2+1}\) = -\(\frac{3x+2}{x^2+1}\) - \(\frac{4\left(x-1\right)}{x^2+1}\)
=> 3x+1 = -3x-2-4x+4
<=>3x+3x+4x = -1-2+4
<=> 10x = 1
<=> x =\(\frac{1}{10}\)(TMĐK)
Bài 1 Giải phương trình
a, \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
b, \(\left(x+2\right)^2+\left(x-1\right)\left(x+3\right)=2\left(x-4\right)\left(x+4\right)\)
c, \(\frac{3}{x-1}=\frac{3x+2}{1-x^2}-\frac{4}{x+1}\)
d, \(\frac{1}{x+1}+\frac{2x^2+1}{x^3+1}+\frac{2x^3-2x^2}{x^2-x+1}=2x\)
Bài 2 : Giải phương trình
\(x^4+3x^3+6x+4=0\)
các bạn ơi ! giúp mik với đi !! mai kiểm tra rồi
bài 1: tính
\(A=\frac{4^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(B=\frac{3x+1}{x^2-2x+1}-\frac{1}{x+1}+\frac{x+3}{1-x^2}\)
\(C=\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x2}\right)\)
\(D=\left(\frac{x^2}{y^2}+\frac{y}{x}\right):\left(\frac{x}{y^2}-\frac{1}{y}+\frac{1}{x}\right)\)
\(E=\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(F=\frac{1}{x-1}-\frac{x^3-x}{x^2+1}\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)
Mấy thánh ơi giúp em với mai phải nộp rồi!!!!!!!!!!!!!!!
Tìm các gtrị của biến số x để phân thức sau bằng 0 :
a) \(\frac{2x-1}{5x-10}\)
b) \(\frac{x^2-x}{2x}\)
c) \(\frac{2x+3}{4x-5}\)
d) \(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}\)
e) \(\frac{x^2-1}{x^2-2x+1}\)
f) \(\frac{x^2-4}{x^2+3x-10}\)
g) \(\frac{x^3-16x}{x^3-3x^2-4x}\)
h) \(\frac{x^3+x^2-x-1}{x^3+2x-3}\)
Cc giúp mk nha !!!!
Mai mk p nộp r
\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
Tìm x
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
giải nhanh hộ mình với, mai mình nộp rồi
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{6}:3\)
\(\Rightarrow x=\frac{1}{18}\)
tim x
a) \(2\frac{1}{2}-x+\frac{4}{5}=\frac{2}{3}-\left(\frac{-4}{7}\right)\)
b) \(\frac{-4}{7}-x=\frac{3}{5}-2x\)
c) \(\frac{9}{2}-\left[\frac{2}{3}-\left(x+\frac{7}{4}\right)\right]=\frac{-5}{4}\)
giúp mình với , mai mình cần rồi
GIẢI PHƯƠNG TRÌNH
a)\(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
b)\(2x^3-5x^2+3x\)
Các pác giúp iem nhaa
Hứa trả ơn full<3
\(a.\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\)\(0\)
\(\Leftrightarrow\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2.\left(x+1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow2x^2-6=0\)
\(\Leftrightarrow2x^2=6\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\sqrt{3}\)
\(b.2x^3-5x^2+3x=0\)
\(\Leftrightarrow x.\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x.\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x.\left[2x.\left(x-1\right)-3.\left(x-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x-1\right).\left(2x-3\right)=0\)
Đến đây tự làm nhé có việc bận
a) \(ĐKXĐ:\hept{\begin{cases}x\ne3\\x\ne-1\end{cases}}\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là :\(S=\left\{0\right\}\)
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
hoặc \(2x-3=0\)
hoặc \(x-1=0\)
\(\Leftrightarrow\)\(x=0\)(tm)
hoặc \(x=\frac{3}{2}\)(tm)
hoặc \(x=1\)(tm)
Vậy tập nghiệm của phương trình là : \(S=\left\{0;\frac{3}{2};1\right\}\)
TÌM GIÁ TRỊ LỚN NHẤT (có thể dùng BĐT côsi)
\(y=\left|x\right|\sqrt{25-x^2}Với-5\le x\le5\)
\(f\left(x\right)=\frac{x}{2}+\sqrt{1-x-2x^2}\)
\(E=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)
TÍNH
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\sqrt{1+\frac{1}{4^2}+\frac{1}{5^2}}+...+\sqrt{1+\frac{1}{2012^2}+\frac{1}{2013^2}}\)
GIÚP EM ĐI Ạ, MAI EM PHẢI KIỂM TRA RỒI