giai pt (x+1)(x+2)(x+5)(x+10)=10x^2
Những câu hỏi liên quan
giai pt : a. x^4/2x^2+1 + 2x^2+1/x^4=2
b.(x/x-1)^2+(x/x+1)^2=10/9
c. x^3+3x^2-10x-24=0
Giai pt
- \(\frac{^{\left(x+2\right)^2}}{2x-3}-1=\frac{x^2+10}{2x-3}\)
- \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
Giai pt
- \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
- \(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
a: \(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow x^2+30x+25=x^2+25x\)
=>5x=-25
hay x=-5(loại)
b: \(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
=>2x+7=10
hay x=3/2
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Giải các pt sau quy về pt bậc hai:
a/(x-6)(x-2)(x+1)(x+3)=7x2
b/4(x+5)(x+6)(x+10)(x+12)=\(3x^2\)
c/\(x^4+x^3-10x^2+x+1=0\)
d/\(x^4-10x^3+26x^2-10x+1=0\)
Giai các pt sau:
\(\frac{1}{x^2+2x}+\frac{1}{x^2+6x+8}+\frac{1}{x^2+10x+24}+\frac{1}{x^2+14x+48}=\frac{4}{105}\)
\(ĐKXĐ:x\ne0;-2;-4;-6;-8\)\(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)
\(\Leftrightarrow\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)
Quy đồng làm nốt
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giai pt (giup toi voi)
1.\(\dfrac{4x}{4x^2-8x+7}+\dfrac{3x}{4x^2-10x+7}=1\)
2.\(\dfrac{2x}{x^2+3x-10}\)+\(\dfrac{5x}{x^2+7x-10}=4\)
Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:
ĐKXĐ: ...
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)
Đặt \(4x+\frac{7}{x}-10=t\)
\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)
\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)
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Giai pt:
\(\left(x+8\right)\sqrt{x+7}=x^2+10x+6\)
giai pt: \(4\sqrt{2x-1}\)= x^2+10x-4
GIẢI PT
\(\sqrt{x^2+10x+25}=4\)
\(\sqrt{x-2}+3=5\)
\(\sqrt{x^2-x+4}-x^2+x-2=0\)
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1}{3}\)
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
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