|x^2-x|-|4x+5|=0
mn giúp mk vs
a. (2x - 1)2 - (4x - 3) (x + 5) = 0
b. x(x - 1) = 3 (x - 1)
c. (x - 1) (3x - 7) = (x - 1) (x + 3)
d. (x - 3)2 + 2x - 6 = 0
mn giải thích từng bước giải giúp mình với, mình bị mất gốc r :'(( sắp tới còn ktra cuối kì nữa
\(a.\left(2x-1\right)^2-\left(4x-3\right)\left(x+5\right)=0\) \(\Leftrightarrow4x^2-4x+1-\left(4x^2+17x-15\right)=0\)
\(\Leftrightarrow-21x+16=0\Leftrightarrow x=\dfrac{16}{21}\) . Vậy ...
b.\(x\left(x-1\right)=3\left(x-1\right)\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) . Vậy ...
c.\(\left(x-1\right)\left(3x-7\right)=\left(x-1\right)\left(x+3\right)\Leftrightarrow\left(x-1\right)\left(3x-7-x-3\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\) . Vậy ...
d.\(\left(x-3\right)^2+2x-6=0\Leftrightarrow\left(x-3\right)\left(x-3+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) . Vậy ...
\(2\sqrt{a}-a\sqrt{\dfrac{4}{a}}\)
Rút gọn với x > 0
Mn giúp e vs ạ ê đang cần gấp
\(2\sqrt{a}-a\sqrt{\dfrac{4}{a}}\)
\(=2\sqrt{a}-a.\dfrac{\sqrt{4}}{\sqrt{a}}\)
\(=2\sqrt{a}-a.\dfrac{2}{\sqrt{a}}\)
\(=2\sqrt{a}-2\sqrt{a}\)
\(=0\)
a)\(x^2-6x+10>0\) vs mọi x
b) \(4x-x^2-5< 0\)vs mọi x
GIÚP MK VS MK ĐG CẦN GẤP
a) Ta có : \(x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1\ge1>0\forall x\)
b) Ta có : \(4x-x^2-5\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)
Vậy ...
câu 1:
a) 4x-5=23 b) |-2x|=5x+14 c) \(\dfrac{x+1}{x-1}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{x^2+2}{x^2-1}\)
mn giúp mk vs, mk cần gấp
Câu 1 :
a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)
b.
|-2x|=5x+14
Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14
<=> - 2x = 5x + 14
<=> - 2x - 5x = 14
<=> - 7x = 14
<=> x = - 2 (thoã mãn)
Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.
Ta có pt: 2x = 5x + 14
<=> - 3x = 14
<=> x = \(-\dfrac{14}{3}\)
Vậy pt có nghiệm x = - 2
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)
\(a,4x-5=23\)
\(\Leftrightarrow4x=23+5\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(b,\left|-2x\right|=5x+14\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)
\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)
\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
a) \(4x-5=23\)
\(4x=23+5\)
\(4x=28\)
\(x=7\)
b) \(\left|-2x\right|=5x+14\)
\(\Leftrightarrow\) \(-2x-5=14\)
\(\Leftrightarrow\) \(-7x=14\)
\(\Leftrightarrow\) \(x=-2\)
\(\Leftrightarrow\) \(-2x=-\left(5x+14\right)\)
\(\Leftrightarrow\) \(-2x=-\left(5x-14\right)\)
\(\Leftrightarrow\) \(-2x+5x=-14\)
\(\Leftrightarrow\) \(3x=-14\)
\(\Leftrightarrow\) \(x=-\dfrac{14}{3}\) \(\left(\text{vô lí}\right)\)
\(\Leftrightarrow x=-2\)
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x-1}+\dfrac{-1}{x+1}=\dfrac{x^2+2}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)+\left(-1\right)\left(x-1\right)=x^2+2\)
\(\Leftrightarrow x^2+x+2=x^2+2\)
\(\Leftrightarrow x+2=2\)
\(\Leftrightarrow x=0\)
2) giải pt
3) \(\sqrt{4x+1}=x+1\)
4) \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
5) \(\sqrt{4x^2-12x+9}=7\)
6) \(5\sqrt{9x-9}-\sqrt{4x-4}-\sqrt{x-1}=36\)
giúp mk vs ah
3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Tìm x
a) |4x/5 - 2/7| -3/2= -2/5
b) | 4x - 1/5|= |2x+ 1/2|
giúp mk vs
\(a,\left|\frac{4x}{5}-\frac{2}{7}\right|-\frac{3}{2}=-\frac{2}{5}\)
\(\Leftrightarrow\left|\frac{4x}{5}-\frac{2}{7}\right|=\frac{11}{10}\)
Xét cả hai trường hợp :
Trường hợp 1 : \(\frac{4x}{5}-\frac{2}{7}=\frac{11}{10}\)
\(\Leftrightarrow\frac{4x}{5}=\frac{97}{70}\)
\(\Leftrightarrow4x=\frac{97}{14}\)
\(\Leftrightarrow x=\frac{97}{56}\)
Trường hợp 2 : \(\frac{4x}{5}-\frac{2}{7}=-\frac{11}{10}\)
\(\Leftrightarrow\frac{4x}{5}=-\frac{57}{50}\)
\(\Leftrightarrow4x=-\frac{57}{14}\)
\(\Leftrightarrow x=-\frac{57}{56}\)
\(b,\left|4x-\frac{1}{5}\right|=\left|2x+\frac{1}{2}\right|\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{5}=2x+\frac{1}{2}\\4x-\frac{1}{5}=-2x+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{5}-2x=\frac{1}{2}\\4x-\frac{1}{5}-(-2x)=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-2x-\frac{1}{5}=\frac{1}{2}\\4x-(-2x)-\frac{1}{5}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{10}\\6x=-\frac{3}{10}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=-\frac{1}{20}\end{cases}}\)
2(x+3)=4x-(2+x)
giúp mk vs mk đag vội
2(x+3)=4x-(2+x)
<=>2x+6=4x-2-x
<=>2x+6-3x+2=0
<=> 8-x=0
<=> x=8
2(x+3)=4x-(2+x)
giúp mk vs mk đag vội
mệt
refer
https://www.google.com/search?q=2(x%2B3)%3D4x-(2%2Bx)&sourceid=chrome&ie=UTF-8
2(x+3)=4x-(2+x)
2x+6=4x-2-x=3x-2
3x-2x=6+2
x=8
2x+6= 4x-2-x
2x-4x+x=-2-6
-x=-4
X=4
Vậy...........
(3/2 x - 1/5)^2 . (x^2 +1/2) = 0
Mn nhớ giải thích chi tiết giúp mình
\(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{5}=0\\x^2+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{1}{5}\\x^2=-\dfrac{1}{2}\left(VLý\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{2}{15}\)