giải pt:
2x^2+√x^2−2x−2=2x+7
Giải pt:
\(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
ĐKXD: ∀x
Ta có \(\dfrac{x^{2^{ }}+2x+1}{x^2+2x+2}\) + \(\dfrac{x^2+2x+2}{x^2+2x+3}\) = \(\dfrac{7}{6}\)
Đặt x2 + 2x + 2 là a (a ∈ Q) Ta có phương trình mới ẩn a:
\(\dfrac{a-1}{a}+\dfrac{a}{a+1}\) = \(\dfrac{7}{6}\)
⇔ \(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}\)+\(\dfrac{6a^2}{6a\left(a+1\right)}\) = \(\dfrac{7}{6}\)
⇔\(\dfrac{6\left(a^2-1\right)+6a^2}{6a\left(a+1\right)}\) = \(\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
Suy ra: 6a2 - 6 + 6a2 = 7a2 + 7a
⇔ 12a2 - 6 - 7a2 - 7a
⇔ 5a2 - 7a - 6 = 0
⇔5a2 - 10a + 3a - 6 = 0
⇔5a( a - 2 ) + 3( a - 2 ) = 0
⇔ (5a + 3)(a - 2) = 0
⇔\(\left[{}\begin{matrix}a-2=0\\5a+3=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}a=2\\a=-0,6\end{matrix}\right.\)
Với a = 2 thì:
x2 + 2x + 2 = 2 ⇔ x2 + 2x = 0
⇔ x(x + 2) = 0 ⇔ \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Với a = -0,6 thì:
x2 + 2x + 2 = -0,6 ⇔ x2 + 2x + 1 = -1,6
⇔ (x + 1)2 = -1,6 (Vô lí vì (x + 1)2 ≥ 0)
Vậy S ∈ \(\left\{0;-2\right\}\)
Giải Pt sau:
(x2+2x+1)/(x2+2x+2)+(x2+2x+2)/(x2+2x+3)=7/6
giải pt ạ
\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-3}+3\right|=14\)
\(\Leftrightarrow2\sqrt{2x-5}=10\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow2x-5=25\)
\(\Leftrightarrow x=15\)
giải pt và bất pt sau:
a.5|2x-1|-3=7
b.(2x+3)(x-2)-x^2+4=0
c. 2x-3/2<1-3x/-5
a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)
TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)
TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
a) Giải pt: \(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)
b)Giải hệ pt \(\left\{{}\begin{matrix}xy-y^2+2y-x-1=\sqrt{y-1}-\sqrt{x}\\3\sqrt{6-y}+3\sqrt{2x+3y-7}=2x+7\end{matrix}\right.\)
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
Giải pt
/x+5/=2x-1
/4x-3/-2x=2x+7
(3x+2) (1-2x)=4x2-1
bạn gải theo 2 TH ở 2 câu đầu còn câu sau nhân ra rồi tính
giải pt 2x-13 /2x-16 + 2x-12/x-8 = 7/8 + 2(5x-39)/3x-24
giải bất pt: \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}< 2x+\dfrac{1}{2x}-7\)
ĐKXĐ: \(x>0\)
\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)
\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)
Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)
\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)
\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)
\(\Rightarrow a-3>0\Rightarrow a>3\)
\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)
\(\Leftrightarrow2x-6\sqrt{x}+1>0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)
giải pt \(6\left(x^2+x+1\right)^2+2x^2+2x-3-\sqrt{4x+5}=0\)
\(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)