\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
và
\(\left|x-3,4\right|+\left|2.6-x\right|=0\)
xét hàm số
f(x)=\(\sqrt[4]{2x}+2\sqrt[4]{6-x}+\sqrt{2x}+2.\sqrt{6-x}\)
D \(\in\left[0;6\right]\)
f'(x)= \(\frac{1}{2.\left(2x\right)^{\frac{3}{4}}}-\frac{1}{2.\left(6-x\right)^{\frac{3}{4}}}+\frac{1}{\sqrt{2x}}-\frac{1}{\sqrt{6-x}}\)
đặt u=\(\left(2x\right)^{\frac{3}{4}}\) \(\left(u\ge0\right)\), v=\(\left(6-x\right)^{\frac{3}{4}}\) \(\left(v\ge0\right)\)
f'(x)= \(\frac{1}{2}.\frac{\left(v^3-u^3\right)}{\left(u.v\right)^3}+\frac{v-u}{u.v}=\frac{\left(v-u\right).\left(v^2+u.v+u^2\right)}{\left(u.v\right)^3}+\frac{v-u}{u.v}=\left(v-u\right).\left(\frac{v^2+u.v+u^2}{\left(u.v\right)^3}+\frac{1}{u.v}\right)\)
\(=\left(v-u\right).g\left(u,v\right)\) ... với g(u,v) > 0
Vậy f'(x) = [(√(2x) - √(6-x)] .G(x), G(x)>0
f'(x)=0 <=> √(2x) - √(6-x) = 0 <=> x=2
lập bảng biến thiên:
tự vẽ
tính f(0), f(2), f(6)
ta được f(x)=m có 2 nghiệm
<=> f(0) \(\le\)m < f(2)
<=> \(2.6^{\frac{1}{4}}+2\sqrt{6}\le m< 3.2^{\frac{1}{4}}+6\)
Cho x và y là hai số khác 0 và thỏa mãn x+y khác 0. Chứng minh rằng:
\(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3y^3}\)
\(0=-\frac{\left(x+2\right)^2+12}{\left(x+2\right)^2}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2}-\frac{\left(x+3\right)^2+3}{\left(x+3\right)^2}+\frac{\left(x+4\right)^2+4}{\left(x+4\right)^2}\)
Tìm x biết
a)\(\frac{x+1}{x-4}>0\)
b)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
c)\(\left(x+2\right)\left(x-3\right)< 0\)
d)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|\le0\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
\(\Rightarrow\frac{x-4}{x-4}+\frac{5}{x-4}>0\)
\(\Rightarrow1+\frac{5}{x-4}>0\)
\(\Rightarrow\frac{5}{x-4}>-1\)
\(\Rightarrow\frac{-5}{-x+4}>-\frac{5}{5}\)
\(\Rightarrow-x+4< -5\)
\(\Rightarrow-x< -9\)
\(\Rightarrow x>9\)
a)\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
b)\(\left(\frac{4}{5}+x\right).\left(x-\frac{8}{13}\right)=0\)
c)\(\left(2x-\frac{1}{2}\right).\left(x-3\right)=0\)
d)\(x+3\frac{1}{2}x+x=\frac{1}{2}\)
a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
\(\frac{-3x.\left(5x+3\right)}{1+3x}>=0\)\(\frac{-2x^2+5x-3}{-x.\left(3x+7\right)}>0\)\(\frac{1}{x-2}-\frac{4}{x^2-4}< \frac{1}{3}\)\(x^2-20x+51>0\)\(\left(x-3\right).\left(2x+1\right)\left(1-5x\right)< 0\)\(\left(x-2\right)\left(x+3\right)=< 0\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(x^2-3x=0\)
đâu phải toán lớp 1
bạn chọn nhầm à
giải pt
a) \(x^2+4x-3\left|x+2\right|+4=0\)
b) \(\left(x+2\right)^2-3\left|x+2\right|-4=0\)
c) \(\left(x^2-3\right)^2-6\left|x^2-3\right|+5=0\)
d) \(\frac{x^2-4x+4}{x^2-2x+1}+\frac{\left|2x-4\right|}{x-1}=3\)
e) \(\left|\frac{2x-1}{x+2}\right|-2\left|\frac{x+2}{2x-1}\right|=1\)
f) \(x^2+\frac{1}{x^2}-10=2\left|x-\frac{1}{x}\right|\)
a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)
b/
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)
\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)
\(\Leftrightarrow\left|x+2\right|-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
c/
\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)
\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)
Đặt \(\frac{\left|x-2\right|}{x-1}=a\)
\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)
e/ ĐKXĐ: ...
Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)
\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)
f/ ĐKXĐ: ...
Đặt \(\left|x-\frac{1}{x}\right|=a\ge0\Rightarrow a^2=x^2+\frac{1}{x^2}-2\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)
Phương trình trở thành:
\(a^2+2-10=2a\)
\(\Leftrightarrow a^2-2a-8=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left|x-\frac{1}{x}\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{x}=4\\x-\frac{1}{x}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-1=0\\x^2+4x-1=0\end{matrix}\right.\)
Tìm x, biết:
a)\(\left(x+5\right).\left(x+9\right)>0\)
b)\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35