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Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
a/ |x-2| = 3
<=> \(\left\{{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b/ |x+1| = |2x + 3|
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x+3\\x+1=-2x-3\\-x-1=2x+3\\-x-1=-2x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\\x=-\dfrac{4}{3}\\x=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
c/ |3x|=x+6
\(\Leftrightarrow\left\{{}\begin{matrix}3x=x+6\\3x=-x-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{3}{2}\end{matrix}\right.\)
a)|x-2|=3
=>\(\left\{{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b)|x+1|=|2x-3|
=>\(\left\{{}\begin{matrix}x+1=2x+3\\x+1=-2x+3\\-x-1=2x+3\\-x-1=-2x-3\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=-2\\x=\frac{-4}{3}\\x=\frac{-4}{3}\\x=-2\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=-2\\x=\frac{-4}{3}\end{matrix}\right.\)
c)|3x|=x+6
=>\(\left\{{}\begin{matrix}3x=x+6\\3x=-x-6\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=3\\x=\frac{-3}{2}\end{matrix}\right.\)
chúc bạn học tốt!
a/ Đặt \(x^2+2x+1=\left(x+1\right)^2=t\ge0\)
\(\Rightarrow\left(t+2\right)t=3\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=1\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b/ \(\Leftrightarrow\left(x^2-x\right)\left(x^2-x+1\right)-6=0\)
Đặt \(x^2-x=t\Rightarrow t\left(t+1\right)-6=0\Rightarrow t^2+t-6=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x=-3\\x^2-x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x+3=0\left(vn\right)\\x^2-x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
ĐKXĐ \(x-1\ne0\) hoặc \(x+3\ne0\)
\(\Rightarrow x\ne1\) và \(x\ne-3\)
\(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)=x^2+3x-x-3-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)
\(\Leftrightarrow9x-x+2x-5x-3x+x=3-5-3-4\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\) (không thỏa ĐK)
Vậy PTVN
b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ: \(x-3\ne0\Rightarrow x\ne3\)
\(x+3\ne0\Rightarrow x\ne-3\)
\(2x+7\ne0\Rightarrow2x\ne-7\Rightarrow x\ne\dfrac{-7}{2}\)
\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow13x+39+x^2+3x-3x-9=12x+42\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\left\{{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\left(KTĐK\right)\\x=-4\left(TĐK\right)\end{matrix}\right.\)
Vậy S={-4}
a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\) ( đk: x ≠ 1 ; x ≠ -3 )
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)
\(\Leftrightarrow3x=-9\)
\(\Rightarrow x=-3\left(KTM\right)\)
S = ∅
b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
( đk: x ≠ ± 3 ; x ≠ \(\dfrac{-7}{2}\) )
\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow13x+39+x^2-9=12x+42\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow x^2+3x-4x-12=0\)
\(\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
S = \(\left\{4\right\}\)
Lời giải:
a)
\((x-2)(x-3)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)(x-2-1)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)^2-(x-2)+2x=(x-2)^2-2\)
\(\Leftrightarrow x+4=0\Rightarrow x=-4\)
b)
\((x-1)^2+3x(x-1)+7=(2x-1)^2+5(x-3)\)
\(\Leftrightarrow (x-1)^2+3x(x-1)+7=x^2+(x-1)^2+2x(x-1)+5(x-3)\)
\(\Leftrightarrow x(x-1)+7=x^2+5(x-3)\)
\(\Leftrightarrow 6x=22\Rightarrow x=\frac{11}{3}\)
c)
\(5(x^2-2x-1)+2(3x-2)=5(x+1)^2=5(x^2-2x+1)\)
\(\Leftrightarrow -5+2(3x-2)=5\)
\(\Leftrightarrow 3x-2=5\Rightarrow x=\frac{7}{3}\)
d)
\((x-1)(x^2+x+1)-2x=x(x-1)(x+1)=x(x^2-1)\)
\(\Leftrightarrow x^3-1-2x=x^3-x\Leftrightarrow -1-x=0\Rightarrow x=-1\)
\(a,\frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}\)
\(\frac{13}{12}x=\frac{13}{12}\Rightarrow x=1\)
a: \(\Leftrightarrow\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{5}x-1\right)^3\)
=>1/3x-1=1/5x-1
=>2/15x=0
hay x=0
b: Đặt 2x+1=a; 3x-1=b
Theo đề, ta có \(\left(a+b\right)^3-a^3-b^3=0\)
=>3ab(a+b)=0
=>5x(2x+1)(3x-1)=0
hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
c: Đặt x-3=a; x+1=b
Theo đề, ta có: \(\left(a+b\right)^3=a^3+b^3\)
=>3ab(a+b)=0
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;-1;1\right\}\)
\(\Leftrightarrow1+3x+3x^2+x^3+1-3x+3x^2-x^3=6\left(x+1\right)^2\)
\(\Leftrightarrow6x^2+2=6\left(x^2+2x+1\right)\)
\(\Leftrightarrow6x^2+2=6x^2+12x+6\)
\(\Leftrightarrow12x=-4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
1+3x+3x2+1+1-3x2+3x-1=6(x2+2x+1)
6x2 +2 = 6x2+12x+6
-12x - 4 = 0
=> x = 3