GIẢI pt
\(A^3_X+C^{x-2}_x=14x\)
giải pt
\(C^1_x+6C^2_x+6C^3_x=9x^2-14x\)
ĐK :\(x\ge3;x\in N\)
áp dụng công thưc tổ hợp ta có
\(\frac{x!}{\left(x-1\right)!}+6\frac{x!}{\left(x-2\right)!2!}+6\frac{x!}{\left(x-3\right)!3!}=9x^2-14\Rightarrow x+3x\left(x-1\right)+x\left(x-1\right)\left(x-2\right)=9x^2-14x\)
suy ra \(x+3x^2-3x+\left(x^2-x\right)\left(x-2\right)-9x^2+14x=0\Rightarrow x\left(17-9x+x^2\right)=0\)
giải pt đối chiếu với đk của x ta tìm đc x
giải pt
\(C^1_x+C^2_x+C^3_x=\frac{7}{2}x\)
đk \(x\ge3;x\in N\)
ÁPdụng công thức tổ hợp ta có
\(\frac{x!}{\left(x-1\right)!}+\frac{x!}{\left(x-2\right)!2}+\frac{x!}{\left(x-3\right)!3!}=\frac{7}{2}x\Rightarrow x+\frac{x\left(x-1\right)}{2}+\frac{x\left(x-1\right)\left(x-2\right)}{6}=\frac{7}{2}x\)
suy ra \(x\left(1+\frac{x-1}{2}+\frac{\left(x-1\right)\left(x-2\right)}{6}-\frac{7}{2}\right)=0\)
giải pt đối chiếu với đk của x ta suy ra đc nghiệm của pt
\(\text{Giải pt}\)
\(A^2_{x-2}+C^{x-2}_x=101\)
Giải :
\(A^2_{x-2}+C^{x-2}_x=101\)\(\left(ĐK:\hept{\begin{cases}x\in Z\\x\ge4\end{cases}}\right)\)
\(\Leftrightarrow\frac{\left(x-2\right)!}{\left(x-4\right)!}+\frac{x!}{\left(x-2\right)!2!}=101\)
\(\Leftrightarrow\left(x-2\right).\left(x-3\right)+\frac{x.\left(x-1\right)}{2}=101\)
\(\Leftrightarrow2.\left(x-2\right).\left(x-3\right)+x.\left(x-1\right)=202\)
\(\Leftrightarrow2x^2-6x-4x+12+x^2-x-202=0\)
\(\Leftrightarrow3x^2-11x-190=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=10\left(tm\right)\\x=\frac{-19}{3}\left(l\right)\end{cases}}\)
\(\frac{2x+2}{4}\)_x=\(\frac{x-2}{x-4}\)giải pt ạ
\(\Leftrightarrow\frac{-x+1}{2}=\frac{x-2}{x-4}\)
\(\Leftrightarrow x^2+4x-3=2x-4\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Bài này là bài lớp 8 mà.
Phương trình tương đương :
\(\frac{2\left(x+1\right)}{4}-x=\frac{x-2}{x-4}\)
<=> \(\frac{x+1}{2}-x=\frac{x-2}{x-4}\)
<=> \(\frac{x\left(x+1\right)}{2}-\frac{x-2}{x-4}=0\)
<=> \(\frac{x\left(x+1\right)\left(x-4\right)}{2\left(x-4\right)}-\frac{2\left(x-2\right)}{2\left(x-4\right)}=0\)
<=> \(\frac{x^3-4x^2+x^2-4x-2x+4}{2\left(x-4\right)}=0\)
<=> \(x^3-4x^2+x^2-4x-2x+4=0\)
<=> ....
giải pt :
a,\(\sqrt{x+14\sqrt{14x-49}}+\sqrt{x-14\sqrt{14x-49}}=\sqrt{14}\)
b, \(\sqrt{x-1+2\sqrt{x-1}}-\sqrt{x-1-2\sqrt{x-1}}=1\)
giải pt :
a,\(3\sqrt{x^2+4x-5}+\sqrt{x-3}=\sqrt{11x^2+25x+2}\)
b,\(\sqrt{5x^2+14x+9}-5\sqrt{x+1}=\sqrt{x^2-x-2}\)
c, \(x^2-8x+17=3\sqrt{x^3-7x+6}\)
giải pt
a) \(\sqrt{x+2\sqrt{x-1}}+3\sqrt{x+8-6\sqrt{x-1}}=1-x\)
b) \(\sqrt{x\sqrt{x-1}-2x+2}+\sqrt{\left(x+3\right)\sqrt{x-1}-4x+4}=\sqrt{x-1}\)
c) \(\sqrt{14x+14\sqrt{14x-49}}+\sqrt{14x-14\sqrt{14x-49}}=14\)
d) \(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}}=4\)
a/ ĐKXĐ: \(x\ge1\)
Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm
b/ \(x\ge1\)
\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)
- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)
- Với \(0\le a\le1\) ta được:
\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)
- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)
c/ ĐKXĐ: \(x\ge\frac{49}{14}\)
\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)
Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(7-\sqrt{14x-49}\ge0\)
\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)
Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)
d/ ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}-2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|-2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=4\)
TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)
\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=4\)
\(\Leftrightarrow\sqrt{2x-1}=5\)
\(\Leftrightarrow x=13\)
TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)
\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)
TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)
\(\sqrt{2x-1}-1-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow4=4\) (luôn đúng)
TH4: \(\frac{1}{2}\le x< 1\)
\(1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(l\right)\)
Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)
Giải pt: \(\sqrt{4x^2-14x+16}+1=x+\sqrt{x^2-4x+5}\)
ak,,,,,,,còn mỗi bước GPT nghiệm nguyên nữa mà mãi ko ra
giải PT:
\(\sqrt[]{3x+1}-\sqrt[]{6-x}+3x^2-14x-8x\)