Tìm X
Xx2+15=1,5x(x+20)
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Ba cặp cắt nhau là :
y= 0,8x +2 và y= \(\dfrac{3}{2}\)x +6
y=0,8x +2 và y= 15-1,5x
y=\(\dfrac{4}{5}\)x -19 và y= \(-\dfrac{3}{2}\)x +6
Các cặp //
y= 15-1.5x//y=\(-\dfrac{3}{2}\)x+6( vì -1,5 = \(-\dfrac{3}{2}\); 15\(\ne\)6)
y=0,8x+2//y=\(\dfrac{4}{5}\)x-19(vì 0,8=\(\dfrac{4}{5}\); 2\(\ne-19\))
chỉ có 2 cặp // nha bn
x+x+2,5x-1,5x=303
2x+2,5x-1,5x=303
4,5x-1,5x=303
3x=303
x=303:3
x=101
Vậy x=101
\(2x+2,5x-1,5x=303\) <=> \(\left(2+2,5-1,5\right)x=303\)<=> \(3x=303\)<=> \(x=101\)
\(1,5x^2-4\left(x^2-2x+1\right)+20=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4+20=0\)
\(\Leftrightarrow x^2+8x+16=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x+4=0\Rightarrow x=-4\)
\(2,x\left(x-2\right)-5x+10=0\)
\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
a) \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x.\left(2x-9\right)-x.3\left(x-5\right)=0\)
\(\Leftrightarrow x.\left[\left(2x-9\right)-3\left(x-5\right)\right]=0\)
\(\Leftrightarrow x.\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x.\left(6-x\right)=0\)
\(\Leftrightarrow S=\left\{0;6\right\}\)
b) \(0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left[0,5x-\left(1,5x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(1-x\right)=0\)
\(+x-3=0\Rightarrow x=3\)
\(+1-x=0\Rightarrow x=1\)
\(\Rightarrow S=\left\{1;3\right\}\)
c) \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow\left(3x-15\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(3-2x\right)\left(x-5\right)=0\)
\(\Rightarrow3-2x=\frac{3}{2}\Rightarrow x-5\Rightarrow x=5\)
\(\Rightarrow S=\left\{5;\frac{3}{2}\right\}\)
a)
\(x\left(2\times-9\right)=3\times\left(\times-5\right)\)
\(\text{⇔}x.\left(2\times-9\right)-x.3\left(x-5\right)=0\)
\(\text{⇔}x.[\left(2\times-9\right)-3\left(x-5\right)]=0\)
\(\text{⇔}x.\left(2x-9-3x+15\right)=0\)
\(\text{⇔}x.\left(6-x\right)=0\)
\(\text{⇔}x=0\) hoặc \(6-x=0+6-x=0\)
\(\text{⇔}x=6\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;6\right\}\) BIẾT MỖI CÂU A :))
a) \(\left(\frac{2}{5}x-2\right)-\left(\frac{3}{2}x+1\right)-\left(-4x-\frac{4}{5}\right)=\)\(\frac{18}{5}\)
\(\frac{2}{5}x-2-\frac{3}{2}x-1-4x+\frac{4}{5}=\frac{18}{5}\)
\(\frac{2}{5}x-\frac{3}{2}x-4x=\frac{18}{5}+2+1-\frac{4}{5}\)
\(\frac{8}{20}x-\frac{30}{20}x-\frac{80}{20}x=\frac{14}{5}+3\)
\(\frac{-51}{10}x=\frac{14}{5}+\frac{15}{5}\)
\(\frac{-51}{10}x=\frac{29}{5}\)
\(x=\frac{29}{5}.\frac{-10}{51}\)
\(x=\frac{-58}{51}\)
vậy \(x=\frac{-58}{51}\)
\(x\times2+15=1,5\times\left(x+20\right)\)
\(x\times2+15=1,5\times x+30\)
\(\Rightarrow x\times2-1,5\times x=30-15\)
\(x\times\left(2-1,5\right)=15\)
\(x\times0,5=15\)
\(x=15:0,5\)
\(x=30\)
\(x\times2+15=1,5\times\left(x+20\right)\)
\(\Rightarrow x\times2+15=1,5\times x+\left(1,5\times20\right)\)
\(\Rightarrow x\times2+15=1,5\times x+30\)
\(\Rightarrow1,5\times x-x\times2=15-30\)
\(\Rightarrow\left(1,5-2\right)\times x=-15\)
\(\Rightarrow-0,5\times x=-15\)
\(\Rightarrow x=-15:-0,5\)
\(\Rightarrow x=30\)
Vậy x = 30
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