Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x2 = 12x + 288
⇔ x2 – 12x – 288 = 0
Có a = 1; b’ = -6; c = -288; Δ’ = b’2 – ac = (-6)2 – 1.(-288) = 324 > 0
Phương trình có hai nghiệm:
Vậy phương trình có hai nghiệm x1 = 24 và x2 = -12.
a/ \(\dfrac{1}{2}.\sqrt{x-1}-\dfrac{3}{2}.\sqrt{9x-9}+24.\sqrt{\dfrac{x-1}{64}}=-17\) ( đkxđ : \(x\ge1\) )
\(\Leftrightarrow\dfrac{1}{2}.\sqrt{x-1}-\dfrac{3}{2}.\sqrt{3^2\left(x-1\right)}+24.\sqrt{\dfrac{x-1}{8^2}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}.\sqrt{x-1}-\dfrac{3.3}{2}.\sqrt{x-1}+\dfrac{24}{8}\sqrt{x-1}=-17\)
\(\Leftrightarrow\) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\left(\sqrt{x-1}\right)\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)=-17\)
\(\Leftrightarrow\sqrt{\left(x-1\right)}.\left(-1\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{-17}{-1}=17\)
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=17^2\)
\(\Leftrightarrow x-1=289\)
\(\Leftrightarrow x=289+1=290\)
vậy x= 290 là nghiệm của phương trình a
b/ \(3x-7\sqrt{x}+4=0\) ( đkxđ : \(x\ge0\) )
\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-4=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{4}{3}\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{9}\\x=1\end{matrix}\right.\)
vậy phương trình có tập nghiệm S=\(\left\{1;\dfrac{16}{9}\right\}\)
c/ \(-5x+7\sqrt{x}+12=0\) ( đkxđ: \(x\ge0\) )
\(\Leftrightarrow-\left(5x+5\sqrt{x}-12\sqrt{x}-12\right)=0\)
\(\Leftrightarrow-\left[5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)\right]\)= 0
\(\Leftrightarrow-\left(5\sqrt{x}-12\right)\left(\sqrt{x}+1\right)=0\)
vì \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1>0\)
\(\Rightarrow5\sqrt{x}-12=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Rightarrow x=\dfrac{144}{25}\)
vậy \(x=\dfrac{144}{25}\) là nghiệm của phương trình c
b:
ĐKXĐ: x>0
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
=>x=1
a) ta có \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=-4x^2-4x+6\)
\(\Leftrightarrow\sqrt{12\left(x+\dfrac{1}{2}\right)^2+16}+\sqrt{20\left(x+\dfrac{1}{2}\right)^2+9}=-\left(2x+1\right)^2+7\)ta có : \(VT\ge\sqrt{16}+\sqrt{9}=7\) và \(VT\le7\)
\(\Rightarrow VT=VP\) \(\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
b) điều kiện \(x>0\)
ta có : \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+4=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\) \(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-2=0\)
\(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\Leftrightarrow\dfrac{x+\sqrt{x}}{\sqrt{x}}=2\Leftrightarrow x+\sqrt{x}=2\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=1\left(N\right)\end{matrix}\right.\)
vậy \(x=1\)
b:
ĐKXĐ: x>0
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
=>x=1
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
a) x2 = 12x + 288 ⇔ x2 - 12x + 288 = 0
∆’ = (-6)2 – 1 . (-288) = 36 + 288 = 324
√∆’ = 18
x1 = 6 + 18 = 24, x2 = 6 – 18 = -12
b)
x2 + 
x = 19
⇔ x2 + 7x – 228 = 0, ∆ = 49 – 4 . (-228) = 49 + 912 = 961 = 312
x1 =
= 12, x2 = 
= -19