a) x^2-2x-3. b) 2x^2+5x-3. c) x^2-4x-5. d) x^2+x-12. Giúp mk vs mk cần gấp. Mk cảm ơn các bạn nhiều lắm !😊🙂
Mn ơi giúp mk với ... giải giúp mk bài toán này trg tối nay thôi nhé... Làm ơn mk đang cần gấp lắm!!!
Tìm x biết,
a) \(9\left(3x-2\right)=x\left(2-3x\right)\)
b) \(\left(2x-1\right)^2-\left(2x+5\right)\left(2x-5\right)=18\)
c) \(5x\left(x-5\right)-2x+10=0\)
d) \(x^2-5=0\)
e) \(x^3+5x^2-4x-20=0\)
f) \(x^3+2\sqrt{2x^2}+2x=0\)
Mn ơi giúp mk với ... cảm ơn rất nhiều...!!!
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
â)\(9\left(3x-2\right)=x\left(2-3x\right)\)
\(\Leftrightarrow27x-18=2x-3x^2\)
\(\Leftrightarrow27x-18-2x+3x^2=0\)
\(\Leftrightarrow3x^2+25x-18=0\)
\(\Leftrightarrow3x^2+27x-2x-18=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x+9=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-9\end{cases}}\)
b)\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x-5\right)=18\)
\(\Leftrightarrow4x^2-4x+1-4x^2+25=18\)
\(\Leftrightarrow26-4x=18\)
\(\Leftrightarrow4x=8\)
\(\Rightarrow x=2\)
c)\(5x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow5x^2-10x-2x+10=0\)
\(\Leftrightarrow5x^2-12x+10=0\)
\(\Leftrightarrow x^2-6x+2=0\)
\(\Leftrightarrow x^2-6x+9-7=0\)
\(\Leftrightarrow\left(x-3\right)^2=7\)
\(\Rightarrow\orbr{\begin{cases}x-3=\sqrt{7}\\x-3=-\sqrt{7}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{7}+3\\x=-\sqrt{7}+3\end{cases}}\)
d)\(x^2-5=0\)
\(\Leftrightarrow x^2=5\)
\(\Rightarrow x=\sqrt{5};-\sqrt{5}\)
e)\(x^3+5x^2-4x-20=0\)
\(\Leftrightarrow x^3-2x^2+7x^2-14x+10x-20=0\)
\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+10\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=-2;2\end{cases}}\)
Giải các phương trình sau: (TM ĐK)
1) \(\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
2) \(\dfrac{2x-1}{5-3x}=2\)
3) \(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
4) \(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
mng giúp mk bài này vs. Cảm ơn bạn nhiều
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
Tìm x
a. x - 15%x = 1/3
b. 4/5x - x - 3/2x + 6/5 = 1/2 - 4/3
Các bạn giúp mk với, mk cần gấp lắm
a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
A = ( 4x - 1 ) × ( 3x + 1 ) - 5x( x - 3 ) - ( x - 4 ) × ( x -3 )
B = ( 5x - 2 ) × ( x + 1 ) - 3x( x^2 - x - 3 ) - 2x(x - 5 ) × ( x - 4 )
Giúp mk vs ạ mình đang cần gấp 😊
thực hiện phép chia và tìm x để số dư bằng 0
a)(x^3-x^2-14x+24):(x^3+x-12)
b)(x^5+4x^3+3x^2-5x+15);(x^3-x+3)
c)(2x^4+2^3+3x^2-5x-20):(x^2+x+4)
d)(2x^4-14x^3+19x^2-20x+9):(x^2-4x+1)
giúp mk gấp vs ah!!!!!!
a) 2|x-2|-3|1-x|=1
b) 3x-|2x+1|=2
c)2.|5x-3|-2x=14
d) |3x-2|+5x=4x-10
e) |x|+|x-2|=3
f) |3x-5|=|x+2|
giúp mk , mk like cho!!!!!!!!!!!
cảm ơn trước hen!!!!
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) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x-1}+\dfrac{-1}{x+1}=\dfrac{x^2+2}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)+\left(-1\right)\left(x-1\right)=x^2+2\)
\(\Leftrightarrow x^2+x+2=x^2+2\)
\(\Leftrightarrow x+2=2\)
\(\Leftrightarrow x=0\)
Tìm GTLN :
a, A=2x-x^2+2
b, B=7x-5x^2-3
c, C=-4x-x^2-7
d, D=6x-4-4x^2
CÁC BẠN GIÚP MK VS MK ĐANG CẦN GẤP !!!
a) A = 2x - x2 + 2
= -x2 + 2x + 2
= -(x2 - 2x + 1 - 1) + 2
= -(x - 1)2 + 3
Ta có: -(x - 1)2 ≤ 0 với ∀x
Nên: -(x - 1)2 + 3 ≤ 3 với ∀x
Dấu "=" xảy ra ⇔ -(x - 1)2 = 0
x - 1 = 0
x = 1
Vậy GTLN của biểu thức A là 3 khi x = 1
Các câu còn lại bạn làm tương tự nhé !
bài 1:
CHO ĐA THỨC A ( x ) = 2x^4 - 3x^3 + 1/2 - 4x . tìm đa thức B (x) và đa thức C(x ), sao cho :
a) A (x) = B( x ) = 4x^5 - 2X^2 -1 ;
B) A (x) - C ( x) = 2 x^3 .
bài 2 :
cho các đa thức :
P (x) = 2x^4-x-2x^3 +1 ;
Q (x) = 5x^2-x^3 + 4x ;
H 9X0 = -2x^4 + x^2 +5 .
tính P(x) + Q (x) + H (x) và P (x) - Q (x) - H (x) .
GIÚP MK VỚI MK ĐAG CẦN GẤP LẮM , GÚP MK VỚI MK XIN CẢM ƠN !