Giải các phương trình sau:
a) 2 5 x − 3 = 4 ; b) 7 2 − 3 5 − 2 x = 9 2
c) 3 x − 1 + 4 = 6 − 3 x − 1 ; d) 3 x + 4 2 − 1 = 1 3
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Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
<=> 6.6 - 0.9x = 2,6 + 0,1x - 4
<=> - 0.9x - 0,1x = -6.6 -1,4
<=> -x = -8
<=> x = 8
Vậy x = 8
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
<=> 3,6 - x - 0,5 = x - 5,5 + x
<=> - x - 3,1 = -5,5
<=> - x = -2.4
<=> x = 2.4
Vậy x = 2.4
\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)
\(< =>2x-7=0< =>x=\dfrac{7}{2}\)
b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)
\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)
\(< =>-25x-11=0\)
\(< =>x=-0,44\)
a.\(x^2-25=8\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)
b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
a) \(15 - 4x = x - 5\)
\( - 4x - x = - 5 - 15\) (chuyển vế)
\( - 5x = - 20\)
\(x = \left( { - 20} \right):\left( { - 5} \right)\) (chia cho một số)
\(x = 4\)
Vậy phương trình có nghiệm \(x = 4\).
b) \(\dfrac{{5x + 2}}{4} + \dfrac{{3x - 2}}{3} = \dfrac{3}{2}\)
\(\dfrac{{\left( {5x + 2} \right).3}}{{4.3}} + \dfrac{{\left( {3x - 2} \right).4}}{{3.4}} = \dfrac{{3.6}}{{2.6}}\) (quy đồng mẫu số)
\(\dfrac{{15x + 6}}{{12}} + \dfrac{{12x - 8}}{{12}} = \dfrac{{18}}{{12}}\)
\(15x + 6 + 12x - 8 = 18\) (chia cả hai vế cho một số)
\(15x + 12x = 18 - 6 + 8\) (chuyển vế)
\(27x = 20\) (rút gọn)
\(x = 20:27\) (chia cả hai vế co một số)
\(x = \dfrac{{20}}{{27}}\)
Vậy phương trình có nghiệm \(x = \dfrac{{20}}{{27}}\).
b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)
a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)
a)\(\Leftrightarrow\)\(x^2-4x-21=0\)
\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)
\(\Leftrightarrow\)\(x(x-7)+3(x-7)=0\)
\(\Leftrightarrow\)\((x-7)(x+3)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)
b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)
\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)
\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)
\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)
Lời giải:
a.
PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)
b.
ĐKXĐ: $x\geq \frac{3}{2}$
PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow \sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5\)
\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)
\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)
\(\Leftrightarrow \sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
\(\dfrac{{5x - 3}}{4} = \dfrac{{x + 2}}{3}\)
\(\dfrac{{\left( {5x - 3} \right).3}}{{4.3}} = \dfrac{{\left( {x + 2} \right).4}}{{3.4}}\)
\(\dfrac{{15x - 9}}{{12}} = \dfrac{{4x + 8}}{{12}}\)
\(15x - 9 = 4x + 8\)
\(15x - 4x = 8 + 9\)
\(11x = 17\)
\(x = 17:11\)
\(x = \dfrac{{17}}{{11}}\)
Vậy phương trình có nghiệm \(x = \dfrac{{17}}{{11}}\).
Mấy bài này nhân chéo là được rồi chứ ta
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}