Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/
\(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+2\left(x^2+2x\right)+1=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Rightarrow x=1\)
b/
\(y^2+2y+1+\left(2^x\right)^2-2.2^x+1=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)
c/
ĐKXĐ: \(x\ne\left\{-2;-4;-6;-8\right\}\)
\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}=x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6}\)
\(\Leftrightarrow\frac{1}{x+2}+\frac{4}{x+8}=\frac{2}{x+4}+\frac{3}{x+6}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{2}{x+4}+\frac{4}{x+8}-\frac{3}{x+6}=0\)
\(\Leftrightarrow\frac{-x}{\left(x+2\right)\left(x+4\right)}+\frac{x}{\left(x+8\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{\left(x+2\right)\left(x+4\right)}=\frac{1}{\left(x+6\right)\left(x+8\right)}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x+2\right)\left(x+4\right)=\left(x+6\right)\left(x+8\right)\)
\(\Leftrightarrow8x=-40\Rightarrow x=-5\)
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`
`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`
`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`
`<=> -55x +20 = 24x-138`
`<=> -55x -24x=-138-20`
`<=>-79x=-158`
`<=> x=2`
Vậy pt có nghiệm `x=2`
`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`
`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2)) = 2/(x(x-2))`
`=> x^2 +2x - x +2 = 2`
`<=> x^2 + x =0`
`<=>x(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)
Vậy pt có nghiệm `x=-1`
`c,2x^3 + 6x^2 =x^2 +3x`
`<=> 2x^3 + 6x^2 -x^2 -3x=0`
`<=> 2x^3 + 5x^2 -3x=0`
`->` Đề có sai ko ạ ?
`d,` \(\left|x-4\right|+3x=5\) `(1)`
Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :
`x-4 = 5-3x`
`<=> x+3x=5+4`
`<=> 4x=9`
`<=> x= 9/4 (t//m)`
Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :
`-(x-4) =5-3x`
`<=> -x +4=5-3x`
`<=> -x+3x=5-4`
`<=> 2x =1`
`<=>x=1/2 ( kt//m)`
Vậy phương trình có nghiệm `x=9/4`
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\left(đkxđ:x\ne-2;-8;-4;-6\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}-1+\dfrac{8}{x+8}-1=\dfrac{4}{x+4}-1+\dfrac{6}{x+6}-1\)
\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\left(-x\right)\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
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}
d)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)=7\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Mà x,y nguyên dương\(\Rightarrow x-y-1< x+y+3\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=1\\x+y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-7\\x+y+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Mạn phép ko chép lại đề :
b) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ \(8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\) ( x # 0)
⇔ \(8\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ ( x + 4)2 = 16
⇔ x2 + 8x + 16 = 16
⇔ x( x + 8) = 0
⇔ x = 0 ( KTM) hoặc : x = - 8 ( TM)
KL.....
ảnh k đc rõ mấy, mong bạn thông cảm :)
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...................