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\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow2160-198x=160x+12\)
\(\Leftrightarrow358x=2148\)
\(\Leftrightarrow x=6\)
Vậy nghiệm của pt x=6
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)
\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)
\(\Leftrightarrow-77x-21=48x-396\)
\(\Leftrightarrow125x=375\)
\(\Leftrightarrow3\)
Vậy nghiệm của pt x=3
Biểu thức \(C = - \frac{2}{3}{x^2} + 7x - 4\) là tam thức bậc hai
Biểu thức A không là tam thức bậc hai vì chứa \(\sqrt x \)
Biểu thức B không là tam thức bậc hai vì chứa \({x^4}\)
Biểu thức D không là tam thức bậc hai vì chứa \({\left( {\frac{1}{x}} \right)^2}\)
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)
<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)
<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)
<=> \(14\left(32x+19\right)+560\)
<=> \(36x+72-35x-140=448x+226+560\)
<=> \(-447x=894\)
<=> x = -2
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{840}-\frac{17}{210}=\frac{8x}{15}+\frac{19}{60}+\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{840}.840-\frac{17}{210}.840=\frac{8x}{15}.840+\frac{19}{60}.840+\frac{2}{3}.840\)
\(\Leftrightarrow x-68=448x+226+560\)
\(\Leftrightarrow x-68=448x+826\)
\(\Leftrightarrow x=448x+826+68\)
\(\Leftrightarrow x=448x+894\)
\(\Leftrightarrow-447x=894\)
=> x = -2
a) \(B=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+5\right)\left(x-1\right)}\right)\)\(:\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{9-3x+x^2+10x+25-\left(x^2-1\right)}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(B=\frac{7x+35}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(B=\frac{7\left(x+5\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)\cdot7\left(x-2\right)}=\frac{x+1}{x-2}\)
b) \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\)
\(\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow x^2-4x+5x-20=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\left(KTM\right)\\x=4\left(TM\right)\end{matrix}\right.\)
Với x = 4 ta có \(B=\frac{4+1}{4-2}=\frac{5}{2}\)
Y