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Bạn ơi bạn học lớp 8 rồi bạn có thể giải giú mình 2 bài toán lớp 7 đang đăng ko. Nếu đc minh cảm ơn nhiều nhé
a/ \(2x-3=5x+2\)
\(\Leftrightarrow5x-2x=-3-2\)
\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy..
b. \(2x\left(x-1\right)=2x+2\)
\(\Leftrightarrow2x^2-4x-2=0\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1+\sqrt{2}\right)\left(x-1-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.\)
Vậy...
c/ ĐKXĐ : \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)
\(\Leftrightarrow2x-16=0\)
\(\Leftrightarrow x=8\)
Vậy..
a: =>4x-2x-2-3x-2=0
=>-x-4=0
=>x=-4
b: =>x+2-2x-2+x=0
=>0x=0(luôn đúng)
d: =>3x=3
hay x=1
e: =>2x=1
hay x=1/2
f: =>4x=-4
hay x=-1
g: =>3x=-3
hay x=-1
c: =>2x+3-5-4+x=0
=>3x-6=0
=>x=2
d: =>3x=3
hay x=1
e: =>2x=1
hay x=1/2
f: =>4x=-4
hay x=-1
g: =>3x=-3
hay x=-1
\(a,4x-2\left(x+1\right)=3x+2\\ \Leftrightarrow4x-2x-2-3x-2=0\\ \Leftrightarrow-x-4=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)
Vậy pt có tập nghiệm \(S=\left\{-4\right\}\)
\(b,x+2-2\left(x+1\right)=-x\\ \Leftrightarrow x+2-2x-2+x=0\\ \Leftrightarrow0=0\)
Vậy pt có tập nghiệm \(S=R\)
\(c,2\left(x+3\right)-5=4-x\\ \Leftrightarrow2x+6-5-4+x=0\\ \Leftrightarrow3x-3=0\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)
Vậy pt có tập nghiệm \(S=\left\{1\right\}\)
\(d,3x-2=1\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)
Vậy pt có tập nghiệm \(S=\left\{1\right\}\)
\(e,2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{2}\right\}\)
\(f,4x+3=-1\\ \Leftrightarrow4x=-4\\ \Leftrightarrow x=-1\)
Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)
\(g,3x+2=-1\\ \Leftrightarrow3x=-3\\ \Leftrightarrow x=-1\)
Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)
\(a,\left(2x-3\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-3+x+1\right)\left(2x-3-x-1\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-4\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\x=4\end{matrix}\right. \\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{2}{3};4\right\}\)
\(b,x^2-6x+9=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2-9\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-3^2\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left[3\left(x-1\right)\right]^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left(3x-3\right)^2=0\\ \Leftrightarrow\left(x-3+3x-3\right)\left(x-3-3x+3\right)=0\\ \Leftrightarrow-2x\left(4x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x=0\\4x-6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\dfrac{3}{2}\right\}\)
Giải pt :
a) \(2x\left(x+5\right)-\left(x-3\right)^2=x^2+6\)
\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)
\(\Leftrightarrow16x-15=0\)
\(\Leftrightarrow x=\frac{15}{16}\)
b) \(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\)
\(\Leftrightarrow2x-18=2x\)
\(\Leftrightarrow-18=0\)( vô lí )
=> x thuộc rỗng
c)d) tương tự
e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)
\(\Leftrightarrow5x-2+9-12x=12-2x-14\)
\(\Leftrightarrow-5x+9=0\)
\(\Leftrightarrow x=\frac{9}{5}\)
f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)
\(\Leftrightarrow8x-4=4x+2-1+2x\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\frac{5}{2}\)
Tìm x :
a) \(3x^3-27x=0\)
\(\Leftrightarrow3x\left(x^2-9\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(2x^3-12x^2+18x=0\)
\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a) 2 (x + 3) (x + 4) - (x - 2)2 > (x - 1)2
<=> 2 (x2 + 4x + 3x + 12) - (x2 - 4x + 4) > x2 - 2x + 1
<=> 2x2 + 8x + 6x + 24 - x2 + 4x - 4 - x2 + 2x - 1 > 0
<=> 20x + 20 > 0
<=> 20 (x + 1) > 0
=> x + 1 > 0
=> x > -1
b) 5x2 - 18x + 19 - (2x - 3)2 > 0
<=> 5x2 - 18x + 19 - (2x2 - 12x + 9) > 0
<=> 5x2 - 18x + 19 - 2x2 + 12x - 9 > 0
<=> 3x2 - 6x + 10 > 0
\(\Delta=b^2-4ac=\left(-6\right)^2-4.3.10=-84< 0\)
Vậy pt vô nghiệm hay\(x\in\varnothing\)
#Học tốt!!!
a) 2( x + 3 )( x + 4 ) - ( x - 2 )2 > ( x - 1 )2
<=> 2( x2 + 7x + 12 ) - ( x2 - 4x + 4 ) > x2 - 2x + 1
<=> 2x2 + 14x + 24 - x2 + 4x - 4 - x2 + 2x - 1 > 0
<=> 20x + 19 > 0
<=> x > -19/20
Vậy ...