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a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
\(A\left(x\right)=5x^2-5x+3=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0,\forall x\)
⇒ pt vô nghiệm
\(B\left(x\right)=4x^2-3x+7=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{16}>0,\forall x\)
⇒ pt vô nghiệm
\(C\left(x\right)=5x^2-11x+6=\left(5x^2-5x\right)-\left(6x-6\right)\)
\(=5x\left(x-1\right)-6\left(x-1\right)=\left(5x-6\right)\left(x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=1\end{matrix}\right.\)
Vậy ...
a, Ta có :
\(A\left(x\right)=5x^2-5x+1+2=0\Leftrightarrow5x^2-6x+3=0\)
\(\Leftrightarrow5\left(x^2-\dfrac{2.3}{5}+\dfrac{9}{25}-\dfrac{9}{25}\right)+3=0\Leftrightarrow5\left(x-\dfrac{3}{5}\right)^2+\dfrac{6}{5}=0\)( vô lí )
vậy đa thức ko có nghiệm
b, \(B\left(x\right)=4x^2-3x+7=0\Leftrightarrow4\left(x^2-\dfrac{2.3}{8}+\dfrac{9}{64}-\dfrac{9}{64}\right)+7=0\)
\(\Leftrightarrow4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{64}=0\)( vô lí )
Vậy đa thức ko có nghiệm
c, \(C\left(x\right)=5x^2-11x+6=0\Leftrightarrow5x^2-6x-5x+6=0\)
\(\Leftrightarrow5x\left(x-1\right)-6\left(x-1\right)=0\Leftrightarrow\left(5x-6\right)\left(x-1\right)=0\Leftrightarrow x=\dfrac{6}{5};x=1\)
\(5x^2-5x=x-1\)
\(\Leftrightarrow5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=1\end{cases}}\)
\(5x^2-5x=x-1\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow5x^2-x-5x+1=0\)
\(\Leftrightarrow x\left(5x-1\right)-\left(5x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=1\end{cases}}\)
\(\frac{3x+2}{5x+7}=\frac{5x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=25x^2+35x-5x-7\)
\(\Leftrightarrow25x^2+30x-7-15x^2-13x-2=0\)
\(\Leftrightarrow10x^2+17x-9=0\)
.............
`5x(x-3)=(x-2)(5x-1)-5`
`\rightarrow 5x^2-15x= [x(5x-1)-2(5x-1)-5]`
`\rightarrow 5x^2-15x=(5x^2-x-10x+2-5)`
`\rightarrow 5x^2-15x=5x^2-11x-3`
`\rightarrow 5x^2-15x-5x^2+11x+3=0`
`\rightarrow -4x+3=0`
`\rightarrow 4x=3`
`\rightarrow x=`\(\dfrac{3}{4}\)
Vậy, `x=`\(\dfrac{3}{4}\)
Còn biến `y` thì mình k thấy bạn nhé!
Cho mk sửa lại từ dòng thứ 6 (tính cả đề)
`\rightarrow -4x+3=0`
`\rightarrow -4x=-3`
`\rightarrow x=-3/-4`
`\rightarrow x=3/4`
Vậy, `x=3/4`
\(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\\ \Leftrightarrow5x^2-15x=5x^2-11x+2-5\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)
=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
B)=>(3x-1).(5x-34)=(40-5x).(25-3x)
=>15x2-102x-5x+34=1000-120x-125x+15x2
=>15x2-107x+34=1000-245x+15x2
=>15x2-15x2-107x+245x=1000-34
=>0-107x+245x=966
=>138x=966
=>x=7
A,=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
\(5x-\frac{1}{3}=3x+\frac{2}{7}=8-\frac{5x}{2}\)
\(\Leftrightarrow5x-3x=\frac{2}{7}+\frac{1}{3}\)
\(\Leftrightarrow2x=\frac{13}{21}\)
\(\Leftrightarrow x=\frac{13}{42}\)
Thử lại:
\(5x-\frac{1}{3}=5\cdot\frac{13}{42}-\frac{1}{3}=\frac{17}{14}\)
\(3x+\frac{2}{7}=3\cdot\frac{13}{42}+\frac{2}{7}=\frac{17}{14}\)
\(8-\frac{5x}{2}=8-5\cdot\frac{13}{42}\div2=\frac{607}{84}\)( vô lý)
Vậy không có giá trị nào của x thoả mãn
\(5^{x+2}+5^{x+1}+5^x=5^x\left(5^2+5+1\right)=5^x.31=19375\)
=>5x=625=>x=4