\(8(x+\dfrac{1}{x} )^{2} \)\(+4(x^{2}+\dfrac{1}{x^{2} } )^{2}\)\(-4 (x^{2}+\dfrac{1}{x^{2}} )(x+\dfrac{1}{x})^{2} \)\(=(x+4)^{2}\)
giúp mik vs ạ cho mik cách giải pt này vs ạ
\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\)
giúp vs ạ mik đng cần gấp
\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\left(x\ne-2;x\ne2\right)\)
\(< =>\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
suy ra
`5x-10-(x^2 +2x-x-2)=12+x^2 -4`
`<=>5x-10-x^2 -2x+x+2-12-x^2 +4=0`
`<=>-x^2 -x^2 +5x-2x+x-10+2+4=0`
`<=>-x^2 +4x-4=0`
`<=>x^2 -4x+4=0`
`<=>(x-2)^2 =0`
`<=>x-2=0`
`<=>x=2(ktmđk)`
vậy phương trình vô nghiệm
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow5\left(x-2\right)-\left(x-1\right)\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow5x-10-\left(x^2+x-2\right)=12+x^2-4\)
\(\Leftrightarrow-x^2+4x-8=x^2+8\)
\(\Leftrightarrow2x^2-4x+16=0\)
\(\Leftrightarrow2\left(x-1\right)^2+14=0\)
Do \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\14>0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow2\left(x-1\right)^2+14>0\)
Vậy phương trình đã cho vô nghiệm
Giúp mik giải câu này z ạ!!
a)\(\dfrac{x-1}{1x-2}+\dfrac{5}{x+2}2=\dfrac{12}{x^2-4}+1\)
b)\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)
c) (3x - 7) (x + 7)=(5 +x) (3 - x)
a,sửa đề : đk x khác -2; 2
\(x^2+x-2+5x-10=12+x^2-4\)
\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)
b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)
c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)
\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)
a. ko hỉu đề lắm :v
b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)
\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)
\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)
\(\Leftrightarrow3x-12+5+5x-105=0\)
\(\Leftrightarrow8x-112=0\)
\(\Leftrightarrow8x=112\)
\(\Leftrightarrow x=14\)
c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)
\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)
\(\Leftrightarrow4x^2+16x-64=0\)
Nghiệm xấu lắm bạn
GIẢI PT:
a) \(\dfrac{x}{x-5}=\dfrac{x-2}{x-6}\)
b) \(\dfrac{2x}{8-x}-\dfrac{2-2x}{4-x}=1\)
e) \(\dfrac{2x}{x+4}-\dfrac{4x}{x^2-16}=0\)
MN GIẢI BÀI NÀY GIÚP E VỚI Ạ. E ĐANG CẦN GẤP Ạ.
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
GIẢI PT :
1) \(\dfrac{x}{x-5}=\dfrac{x-2}{x-6}\)
2) \(\dfrac{2x}{8-x}-\dfrac{2-2x}{4-x}=1\)
3) \(\dfrac{2x}{x+4}-\dfrac{4x}{x^2-16}=0\)
GIẢI PHƯƠNG TRÌNH VÀ GHI RÕ ĐIỀU KIỆN CỦA CÁC CÂU.
MN GIÚP E BÀI NÀY VỚI Ạ. E ĐANG CẦN GẤP Ạ.
1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)
hay x=10
GIẢI PT :
1) \(\dfrac{x}{x-5}=\dfrac{x-2}{x-6}\) (GHI RÕ ĐK)
2) \(\dfrac{2x}{8-x}-\dfrac{2-2x}{4-x}=1\) (GHI RÕ ĐK)
3) \(\dfrac{2x}{x+4}-\dfrac{4x}{x^2-16}=0\) (GHI RÕ ĐK)
MN GIÚP E BÀI NÀY VỚI Ạ. E ĐANG CẦN GẤP Ạ.
1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)
hay x=10
Tính:
\(\dfrac{x^3+8}{x^2-2x+1}\) X \(\dfrac{x^2+3x+2}{1-x^2}\)
Mn giúp mik vs ạ
Lời giải:
$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$
$=\frac{(x+2)(x^2-2x+4)(x+1)(x+2)}{(x-1)^2(1-x)(x+1)}$
$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
a)(\(\dfrac{31}{35}\)-\(\dfrac{4}{7}\)) x \(\dfrac{8}{7}\) : 2
b)(1 - \(\dfrac{1}{2}\)) x ( 1- \(\dfrac{1}{3}\)) x (1-\(\dfrac{1}{4}\)) x ( 1-\(\dfrac{1}{5}\))
nhanh giúp mik với!!
lời giải chi tiết ạ!!!
\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2
= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2
= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)
= \(\dfrac{44}{245}\)
b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))
= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)
= \(\dfrac{1}{5}\)
Giair phương trình
\(8\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=\left(x+4\right)^2+4\left(x+\dfrac{1}{x}\right)^2\left(x^2+\dfrac{1}{x^2}\right).\)
giúp mik vs mn ơi
Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?
Giúp mik bài này với ạ:
\(\dfrac{x+1}{x-2}\) = \(\dfrac{2}{x^2-4}\)
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)
\(\Rightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}=\dfrac{2}{x^2-4}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=2\)
\(\Leftrightarrow x^2+3x+2=2\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) (thỏa mãn)
đkxđ: \(x ≠2; x ≠-2\)
\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)
\(⇔\dfrac{(x+1)(x+2)}{x^2-4}=\dfrac{2}{x^2-4}\)
\(⇔(x+1)(x+2)=2\)
\(⇔x^2+3x=0\)
\(⇔x(x+3)=0\)
\(⇔\left[\begin{array}{} x=0\\ x+3=0 \end{array} \right.\)
\(⇔\left[\begin{array}{} x=0\\ x=-3 \end{array} \right.\)