Gải phương trình Sau
\(\dfrac{x^2+4x+6}{\text{x}+2}+\dfrac{x^2+16x+72}{x+8}\)=\(\dfrac{x^2+8x+20}{\text{x}+4}+\dfrac{x^2+12x+42}{x+6}\)
giải phương trình \(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\left(đkxđ:x\ne-2;-8;-4;-6\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}-1+\dfrac{8}{x+8}-1=\dfrac{4}{x+4}-1+\dfrac{6}{x+6}-1\)
\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\left(-x\right)\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Giải phương trình :))
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)
ĐKXĐ: x\(\ne\) -2; x\(\ne\) -4; x\(\ne\) -6; x\(\ne\) -8;
\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\) \(\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow\left(x+2+\dfrac{2}{x+2}\right)+\left(x+8+\dfrac{8}{x+8}\right)=\)
\(\left(x+4+\dfrac{4}{x+4}\right)+\left(x+6+\dfrac{6}{x+6}\right)\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
=> 2.(x+4)(x+8)(x+6) + 8(x+2)(x+4)(x+6)=4(x+2)(x+6)(x+8)
+ 6(x+2)(x+4)(x+8)
<=>(2x+8)(x2 + 14x+64) + (8x+48)(x2+6x+8) - (4x+8)(x2 + 14x+64)
-(6x+48)(x2+6x+8)
<=> (x2 + 14x+64)(2x+8 -4x -8) + (x2+6x+8)(8x+48+6x-48)=0
<=> -2x(x2 + 14x+64)+ 2x(x2+6x+8) = 0
<=> -2x3 -28x2 -128x+ 2x3 +12x2 +16x = 0
<=> -16x2 - 112x = 0
<=> -x(16x+112) = 0
<=>\(\left[{}\begin{matrix}x=0\\16x+112=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=7\left(tmđk\right)\end{matrix}\right.\)
vậy S={0;7}
sửa bài:
<=>﴾2x+8﴿﴾x2 + 14x+48﴿ + ﴾8x+48﴿﴾x2 +6x+8﴿ ‐ ﴾4x+8﴿﴾x2 + 14x+48﴿
‐﴾6x+48﴿﴾x2 +6x+8﴿
<=> ﴾x2 + 14x+48﴿﴾2x+8 ‐4x ‐8﴿ + ﴾x2 +6x+8﴿﴾8x+48+6x‐48﴿=0
<=> ‐2x﴾x2 + 14x+48﴿+ 2x﴾x2 +6x+8﴿ = 0
<=> ‐2x3 ‐28x2 ‐96x+ 2x3 +12x2 +16x = 0
<=> ‐16x2 ‐ 80x = 0
<=> ‐x﴾16x+80﴿ = 0
<=>\(\left[{}\begin{matrix}x=0\\16x+80=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
vậy : S={0;-5}
Giải phương trình tìm x, biết:
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\).
\(pt\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Rightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Rightarrow2x+10+\dfrac{2}{x+2}+\dfrac{8}{x+8}=2x+10+\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Rightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Rightarrow\dfrac{2x+16+8x+16}{\left(x+2\right)\left(x+8\right)}=\dfrac{4x+24+6x+24}{\left(x+4\right)\left(x+6\right)}\)
\(\Rightarrow\dfrac{\left(x+2\right)\left(x+8\right)}{2\left(5x+16\right)}=\dfrac{\left(x+4\right)\left(x+6\right)}{2\left(5x+24\right)}\)
\(\Rightarrow\dfrac{x^2+8x+2x+16}{5x+16}=\dfrac{x^2+6x+4x+24}{5x+24}\)
\(\Rightarrow\dfrac{x^2+10x+16}{5x+16}=\dfrac{x^2+10x+24}{5x+24}=\dfrac{x^2+10x+24-x^2-10x-16}{5x+24-5x-16}=\dfrac{8}{8}=1\)(dãy ts bằng nhau)
Dễ dàng tìm được x
Giải pt:
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)
\(\Leftrightarrow1+\dfrac{2}{x+2}+1+\dfrac{8}{x+8}=1+\dfrac{4}{x+4}+1+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{4}{x+8}=\dfrac{2}{x+4}+\dfrac{3}{x+6}\)
\(\Leftrightarrow\dfrac{4}{x+8}-\dfrac{3}{x+6}=\dfrac{2}{x+4}-\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{4x+24-3\left(x+8\right)}{\left(x+8\right)\left(x+6\right)}=\dfrac{2x+4-\left(x+4\right)}{\left(x+4\right)\left(x+2\right)}\)
\(\dfrac{x}{\left(x+8\right)\left(x+6\right)}=\dfrac{x}{\left(x+4\right)\left(x+2\right)}\)
x=0 là nghiệm
x khác 0
\(\left\{{}\begin{matrix}x\ne\left\{-8;-6;-4;-2\right\}\\\left(x+4\right)\left(x+2\right)=\left(x+8\right)\left(x+6\right)\end{matrix}\right.\)<=>x^2 +6x+8 =x^2 +14x+48
-40 =8x=> x =-5 nhận
x={-5;0}
Giải các phương trình sau:
a) \(x\left(x+2\right)\left(x^2+2x+2\right)+1=0\)
b) \(y^2+4^x+2y-2^{x+1}+2=0\)
c) \(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)
Câu a và câu c bn kia làm rồi nên mk làm câu b thôi nhé....
b) y2 + 4x + 2y - 2x+1 + 2 = 0
\(\Leftrightarrow\) (y2 + 2y + 1) + 4x - 2x.2 + 1 = 0
\(\Leftrightarrow\) (y + 1)2 + [(2x)2 - 2.2x.1 + 1] = 0
\(\Leftrightarrow\) (y + 1)2 + (2x - 1)2 = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\2^x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)
Vậy...................
Giải phương trình:
a) \(\dfrac{3x-2}{x^2-12x+20}-\dfrac{4x+3}{x^2+6x-16}=\dfrac{7x+11}{x^2-2x-80}\)
b) \(\dfrac{2x-5}{x^2+5x-36}-\dfrac{x-6}{x^2+3x-28}=\dfrac{x+8}{x^2+16x+63}\)
a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)
=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)
=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22
=>-x^2+59x+14-8x^2+5x+22=0
=>-9x^2+54x+36=0
=>x^2-6x-4=0
=>\(x=3\pm\sqrt{13}\)
b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)
=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)
=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32
=>x^2+6x+19=x^2+4x-32
=>2x=-51
=>x=-51/2
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
Với giá trị nào của x, giá trị của biểu thức sau bagfw 0:
\(\dfrac{1+8x}{4+8x}\) - \(\dfrac{4x}{12x-6}\) + \(\dfrac{\dfrac{32}{3}x^2}{4-16x^2}\) ??
Giải phương trình:
\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)
=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)
=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0
=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)
Voi \(\frac{1}{x+2}-....\)=0 ta co
Dat x+5=t
=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0
=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)
=>t=0
=>x=-5
Vay phuong trinh co nghiem x=0;-5
toán lớp 8 mà đi giải phương trình hả má
ĐKXĐ:\(x\ne-2;-4;-6;-8\)
\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)
\(\frac{10+32}{\left(x+2\right)\left(x+8\right)}=\frac{10x+48}{\left(x+4\right)\left(x+6\right)}\)
\(\Leftrightarrow\frac{5x+16}{\left(x+2\right)\left(x+8\right)}=\frac{5x+24}{\left(x+4\right)\left(x+6\right)}\)
\(\Leftrightarrow\left(5x+16\right)\left(x+4\right)\left(x+6\right)=\left(5x+24\right)\left(x+2\right)\left(x+8\right)\)
\(\Leftrightarrow x^2+5x=0\)(bạn tự biến đổi nhé)
\(\Leftrightarrow x=0;-5\)(tm ĐKXĐ)
Vậy phương trình có nghiệm 0;-5 (mình làm hơi tắt bn thông cảm nha)