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a.
\(\dfrac{5x-17}{14}+\dfrac{x-3}{26}>\dfrac{29-9x}{91}\)
\(\Leftrightarrow13\left(5x-17\right)+7\left(x-3\right)>2\left(29-9x\right)\)
\(\Leftrightarrow65x-221+7x-21>58-18x\)
\(\Leftrightarrow65x+7x+18x>58+21+221\)
\(\Leftrightarrow90x>300\)
\(\Leftrightarrow x>\dfrac{10}{3}\)
b)
\(\dfrac{8x-1}{9}+\dfrac{3x-2}{4}< \dfrac{43+8x}{12}+\dfrac{35x}{36}\)
\(\Leftrightarrow4\left(8x-1\right)+9\left(3x-2\right)< 3\left(43+8x\right)+35x\)
\(\Leftrightarrow32x-4+27x-18< 129+24x+35x\)
\(\Leftrightarrow32x+27x-24x-35x< 129+18+4\)
\(\Leftrightarrow0x< 151\) ( luôn đúng)
Vậy bất pt vô số nghiệm
b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)
\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)
=>-13x=-21
hay x=21/13
c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)
=>x-100=0
hay x=100
\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)
\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)
\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)
\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)
mình ra là \(\dfrac{11}{36}\)mà bn
bn coi lại đi
I=3x2-5x+3
I=3(x2-\(\dfrac{5}{3}\)x+1)
I=3[x2-2.x.\(\dfrac{5}{3}\)+\(\left(\dfrac{5}{6}\right)^2\)-\(\left(\dfrac{5}{6}\right)^2\)+1]
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)≥\(\dfrac{11}{36}\)
vậy Min I= \(\dfrac{11}{36}\)khi x =\(\dfrac{5}{3}\)
Theo mik nghĩ là vậy á
CHÚC BN HỌC TỐT
a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
\(\dfrac{x-1}{9}+\dfrac{x-2}{8}+\dfrac{x-3}{7}=\dfrac{x-9}{1}+\dfrac{x-8}{2}+\dfrac{x-7}{3}\\ \Leftrightarrow\dfrac{x-1}{9}-1+\dfrac{x-2}{8}-1+\dfrac{x-3}{7}-1=\dfrac{x-9}{1}-1+\dfrac{x-8}{2}-1+\dfrac{x-7}{3}-1\\ \Leftrightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-10}{7}=\dfrac{x-10}{1}+\dfrac{x-10}{2}+\dfrac{x-10}{3}\\ \Leftrightarrow\left(x-10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{7}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\Leftrightarrow x-10=0\\ \Leftrightarrow x=10\)
Trừ 2 vế với 1:
\(\Rightarrow\dfrac{x-1}{9}+\dfrac{x-2}{8}+\dfrac{x-3}{7}+3=\dfrac{x-9}{1}+\dfrac{x-8}{2}+\dfrac{x-7}{3}+3\)
\(\Rightarrow\left(\dfrac{x-1}{9}-1\right)+\left(\dfrac{x-2}{8}-1\right)+\left(\dfrac{x-3}{7}-1\right)=\left(\dfrac{x-9}{1}-1\right)+\left(\dfrac{x-8}{2}-1\right)+\left(\dfrac{x-7}{3}-1\right)\)
\(\Rightarrow\left(\dfrac{x-1}{9}-\dfrac{9}{9}\right)+\left(\dfrac{x-2}{8}-\dfrac{8}{8}\right)+\left(\dfrac{x-3}{7}-\dfrac{7}{7}\right)=\left(\dfrac{x-9}{1}-\dfrac{1}{1}\right)+\left(\dfrac{x-8}{2}-\dfrac{2}{2}\right)+\left(\dfrac{x-7}{3}-\dfrac{3}{3}\right)\)
\(\Rightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-3}{7}=\dfrac{x-10}{1}+\dfrac{x-10}{2}+\dfrac{x-10}{3}\)
\(\Rightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-10}{7}-\dfrac{x-10}{1}-\dfrac{x-10}{2}-\dfrac{x-10}{3}\)
\(\Rightarrow\left(x-10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{7}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left(x-10\right)=0\)
\(\Rightarrow x=10\)
a) \(x^4+x^2-2=0\)
\(\Leftrightarrow x^4+2x^2-x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2+2\right)-\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+2=0\) hoặc \(x+1=0\) hoặc \(x-1=0\)
. \(x^2+2=0\Leftrightarrow x^2=-2\) (vô nghiệm)
.. \(x+1=0\Leftrightarrow x=-1\)
... \(x-1=0\Leftrightarrow x=1\)
Vậy \(S=\left\{\pm1\right\}\)
b) \(x^4-13x^2+36=0\)
\(\Leftrightarrow x^4-9x^2-4x^2+36=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-4\left(x^2-9\right)=0
\)
\(\Leftrightarrow\left(x^2-9\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x+3=0\) hoặc \(x-3=0\) hoặc \(x+2=0\) hoặc \(x-2=0\)
. \(x+3=0\Leftrightarrow x=-3\)
.. \(x-3=0\Leftrightarrow x=3\)
... \(x+2=0\Leftrightarrow x=-2\)
.... \(x-2=0\Leftrightarrow x=2\)
Vậy \(S=\left\{\pm3;\pm2\right\}\)
Câu C bạn ghi ko rõ lém!!!!!!!!
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
\(a,\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2.\left(x-1\right)}+\dfrac{-2x}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)}{2.\left(x-1\right).\left(x+1\right)}+\dfrac{\left(-2x\right).x}{x.\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)-2x^2}{x.\left(x+1\right)\left(x-1\right)}\)
b: \(=\dfrac{y^2-12y+24}{6y\left(y-6\right)}\)
c: \(=\dfrac{12-2x+3x}{2x\left(x+3\right)}=\dfrac{x+12}{2x\left(x+3\right)}\)
\(\dfrac{36}{x}+\dfrac{36}{x-12}=\dfrac{9}{2}\)
\(\Rightarrow72\left(x-12\right)+72x\left(x-12\right)-x\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(72+72x-x\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(72+71x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\72+71x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=\dfrac{-72}{71}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 12 hoặc x = \(\dfrac{-72}{71}\) .
sai rồi bạn