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\(x\ne2\)
Áp dụng HĐT \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
\(\left(\frac{x-3}{x-2}\right)^3-\left(x-3\right)^3=16\)
\(\Leftrightarrow\left(\frac{\left(x-3\right)-\left(x-3\right)\left(x-2\right)}{x-2}\right)^3+\frac{3\left(x-3\right)^2}{\left(x-2\right)}\left(\frac{x-3}{x-2}-x+3\right)=16\)
\(\Leftrightarrow\left(\frac{\left(x-3\right)\left(3-x\right)}{\left(x-2\right)}\right)^3+\frac{3\left(x-3\right)^2}{x-2}\left(\frac{\left(x-3\right)\left(3-x\right)}{x-2}\right)=16\)
\(\Leftrightarrow\left(-\frac{\left(x-3\right)^2}{x-2}\right)^3-3.\left(\frac{\left(x-3\right)^2}{x-2}\right)^2=16\)
Đặt \(\frac{\left(x-3\right)^2}{x-2}=a\)
\(-a^3-3a^2=16\Leftrightarrow a^3+3a^2+16=0\Rightarrow a=-4\)
\(\Rightarrow\frac{\left(x-3\right)^2}{x-2}=-4\Leftrightarrow x^2-2x+1=0\Rightarrow x=1\)
Lời giải:
Ta có:
\((x+3)(x+12)(x-4)(x-16)+20x^2=0\)
\(\Leftrightarrow [(x+3)(x-16)][(x+12)(x-4)]+20x^2=0\)
\(\Leftrightarrow (x^2-13x-48)(x^2+8x-48)+20x^2=0\)
Đặt \(x^2-12x-48=a\). PT trở thành:
\((a-x)(a+20x)+20x^2=0\)
\(\Leftrightarrow a^2+19ax-20x^2+20x^2=0\Leftrightarrow a^2+19ax=0\)
\(\Leftrightarrow a(a+19x)=0\)
\(\Leftrightarrow (x^2-12x-48)(x^2+7x-48)=0\)
\(\Leftrightarrow \left[\begin{matrix} x^2-12x-48=0\\ x^2+7x-48=0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=6\pm 2\sqrt{21}\\ x=\frac{-7\pm \sqrt{241}}{2}\end{matrix}\right.\)
Vậy......
b: \(\Leftrightarrow\left(x^2-2x+1-1\right)^2-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left[\left(x-1\right)^2-1\right]^2-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^4-2\left(x-1\right)^2+1-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-3\right)\left(x+1\right)=0\)
hay \(x\in\left\{1;3;-1\right\}\)
a: \(\Leftrightarrow2x^3-3x-10=-2\left(8-12x+6x^2-x^3\right)\)
\(\Leftrightarrow2x^3-3x-10=-16+24x-12x^2+2x^3\)
\(\Leftrightarrow-3x-10+16-24x+12x^2=0\)
=>\(12x^2-27x+6=0\)
hay \(x\in\left\{2;\dfrac{1}{4}\right\}\)
Câu 3: 9x + 5y + 18 = 2xy
<=> 9(x - 2) - 2y(x - 2) = -y - 36
<=> (x - 2)(9 - 2y) = -y - 36
<=> x - 2 = \(\dfrac{-y-36}{9-2y}\) (1)
Do x - 2 nguyên nên \(-y-36⋮9-2y\)
\(\Rightarrow2y+72⋮9-2y\)\(\Rightarrow2y+72+9-2y⋮9-2y\)
\(\Rightarrow81⋮9-2y\)\(\Rightarrow9-2y\in\left\{1;-1;3;-3;9;-9;27;-27;81;-81\right\}\)
\(\Rightarrow y\in\left\{4;5;3;6;0;9;-9;18;-36;45\right\}\)
Thay lần lượt giá trị của y vào (1) ta được các cặp giá trị (x;y) thỏa mãn là: (43;5); (-11;3); (7;9); (1;-9); (3;45)
Câu 4:
a) 2x2 + 2x + 1 = \(\sqrt{4x+1}\) (đk: \(x\ge-\dfrac{1}{4}\))
\(\Rightarrow\left(2x^2+2x+1\right)^2=4x+1\)
<=> 4x4 + 4x2 + 1 + 8x3 + 4x + 4x2 - 4x - 1 = 0
<=> 4x4 + 8x3 + 8x2 = 0 (*)
+) x = 0, thay vào (*) thỏa mãn
+) x \(\ne0\), chia cả 2 vế của (*) cho 4x2 ta được:
x2 + 2x + 2 = 0
<=> (x + 1)2 + 1 = 0, vô nghiệm
Vậy pt có nghiệm x = 0
a) \(\dfrac{12}{x-1}-\dfrac{8}{x+1}=1\) \(\Leftrightarrow\) \(\dfrac{12\left(x+1\right)-8\left(x-1\right)}{x^2-1}=1\)
\(\Leftrightarrow\) \(\dfrac{12x+12-8x+8}{x^2-1}=1\) \(\Leftrightarrow\) \(\dfrac{4x+20}{x^2-1}=1\)
\(\Leftrightarrow\) \(x^2-1=4x+20\) \(\Leftrightarrow\) \(x^2-4x-21=0\)
giải pt ta có 2 nghiệm : \(x_1=7;x_2=-3\)
vậy phương trình có 2 nghiệm \(x=7;x=-3\)
b) \(\dfrac{16}{x-3}+\dfrac{30}{1-x}=3\) \(\Leftrightarrow\) \(\dfrac{16\left(1-x\right)+30\left(x-3\right)}{\left(x-3\right)\left(1-x\right)}=3\)
\(\Leftrightarrow\) \(\dfrac{16-16x+30x-90}{x-x^2-3+3x}=3\) \(\Leftrightarrow\) \(\dfrac{14x-74}{-x^2+4x-3}=3\)
\(\Leftrightarrow\) \(3\left(-x^2+4x-3\right)=14x-74\)
\(\Leftrightarrow\) \(-3x^2+12x-9=14x-74\)
\(\Leftrightarrow\) \(3x^2-2x-65=0\)
giải pt ta có 2 nghiệm : \(x_1=5;x_2=\dfrac{-13}{3}\)
vậy phương trình có 2 nghiệm \(x=5;x=\dfrac{-13}{3}\)
c) ĐK: x\(\ne3,x\ne-2\)
\(\dfrac{x^2-3x+5}{\left(x-3\right)\left(x+2\right)}=\dfrac{1}{x-3}\Leftrightarrow\dfrac{x^2-3x+5}{\left(x-3\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}\Leftrightarrow x^2-3x+5=x+2\Leftrightarrow x^2-4x+3=0\Leftrightarrow x^2-x-3x+3=0\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy S={1}
d) ĐK: \(x\ne2,x\ne-4\)
\(\dfrac{2x}{x-2}-\dfrac{x}{x+4}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x\left(x+4\right)}{\left(x-2\right)\left(x+4\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x^2+8x}{\left(x-2\right)\left(x+4\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x^2+8x-x^2+2x}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow x^2+10x=8x+8\Leftrightarrow x^2+2x-8=0\Leftrightarrow x^2-2x+4x-8=0\Leftrightarrow x\left(x-2\right)+4\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-4\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm
Link tham khảo: https://diendantoanhoc.net/topic/134563-gi%E1%BA%A3i-ph%C6%B0%C6%A1ng-tr%C3%ACnh-fracx-3x-23-x-3316/