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\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)
\(\Leftrightarrow2x^2-5>2x^2-10x-3\)
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)
\(\Leftrightarrow x^2-10x+17-x-7>0\)
\(\Leftrightarrow x^2-11x+10>0\)
=>x>10 hoặc x<1
a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)
⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3
⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7
⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0
⇔x2−11x+10>0⇔x2−11x+10>0
=>x>10 hoặc x<1
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a)
\((x-3)(x-5)(x-6)(x-10)=24x^2\)
\(\Leftrightarrow [(x-3)(x-10)][(x-5)(x-6)]=24x^2\)
\(\Leftrightarrow (x^2-13x+30)(x^2-11x+30)=24x^2\)
Đặt \(x^2-11x+30=a\). PT trở thành:
\((a-2x)a=24x^2\)
\(\Leftrightarrow a^2-2ax-24x^2=0\)
\(\Leftrightarrow a^2-6ax+4ax-24x^2=0\)
\(\Leftrightarrow a(a-6x)+4x(a-6x)=0\)
\(\Leftrightarrow (a+4x)(a-6x)=0\)
\(\Rightarrow \left[\begin{matrix} a+4x=0\\ a-6x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2-7x+30=0\\ x^2-17x+30=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} (x-3,5)^2+17,75=0(\text{vô lý})\\ (x-15)(x-2)=0\end{matrix}\right.\)
\(\Rightarrow x=15\) hoặc $x=2$
b)
Đặt \(x-7=a\). PT trở thành:
\((a+1)^4+(a-1)^4=272\)
\(\Leftrightarrow a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=272\)
\(\Leftrightarrow 2a^4+12a^2+2=272\)
\(\Leftrightarrow a^4+6a^2-135=0\)
\(\Leftrightarrow (a^2+3)^2-144=0\Leftrightarrow (a^2+3)^2-12^2=0\)
\(\Leftrightarrow (a^2+15)(a^2-9)=0\)
\(\Rightarrow a^2-9=0\Rightarrow a=\pm 3\)
\(\Rightarrow x=a+7=\left[\begin{matrix} 4\\ 10\end{matrix}\right.\)
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
a, \(\Leftrightarrow\left(x+1+x-2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)+\left(x-2\right)^2\right]-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-\left(2x-1\right)^2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+7-4x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(-3x^2+3x+6\right)=0\)
\(\Leftrightarrow-3\left(2x-1\right)\left(x^2-x-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)\left(x-2\right)=0\)
=>x=1/2 hoặc x=-1 hoặc x=2
Vậy pt có tập nghiệm là S={1/2;-1;2}
b, \(x^4=24x+32\Leftrightarrow x^4-24x-32=0\)
\(\Leftrightarrow x^4-2x^3-4x^2+2x^3-4x^2-8x+8x^2-16x-32=0\)
\(\Leftrightarrow x^2\left(x^2-2x-4\right)+2x\left(x^2-2x-4\right)+8\left(x^2-2x-4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)
\(\Leftrightarrow x^2-2x-4=0\) (vì x^2+2x+8 > 0)
\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x-1=\pm\sqrt{5}\Leftrightarrow x=1\pm\sqrt{5}\)
Vậy...
c, \(\left(x-6\right)^4+\left(x-8\right)^4=16\)
Đặt x-6=t => x-8=t-2
Ta có: \(t^4+\left(t-2\right)^4=16\Leftrightarrow t^4+t^4-8t^3+24t^2-32t+16=16\)
\(\Leftrightarrow2t^4-8t^3+24t^2-32t=0\Leftrightarrow t^4-4t^3+12t^2-16t=0\)
\(\Leftrightarrow t^4-2t^3-2t^3+4t^2+8t^2-16t=0\)
\(\Leftrightarrow t^3\left(t-2\right)-2t^2\left(t-2\right)+8t\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t^3-2t^2+8t\right)=0\Leftrightarrow\left(t-2\right)t\left(t^2-2t+8\right)=0\)
Mà t^2-2t+8=(t-1)^2+7 > 0
\(\Rightarrow\orbr{\begin{cases}t-2=0\\t=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-6-2=0\\x-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)
Vậy...