Tìm x, biết:
a) \(\left|3x-2\right|-x>1\)
b) \(\left|3x-5\right|+\left|2x+3\right|=7\)
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)
\(\Leftrightarrow42x-39=4\)
\(\Leftrightarrow42x=4+39\)
\(\Leftrightarrow42x=43\)
\(\Rightarrow x=\dfrac{43}{42}\)
Tick cho mk nha
tìm \(x\) biết:
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
hay x=-2
Bài 11 : Tìm GTNN của của các biểu thức sau :
a ) \(A=\left|x+3\right|+\left|2x-5\right|+\left|x-7\right|.\)
b ) \(B=\left|x+2\right|+\left|3x-4\right|+\left|x-2\right|+5\)
c ) \(M=\left|x+2\right|+\left|x-3\right|\)
d ) \(C=\left|2x+5\right|+\left|2x+1\right|+\left|2x-7\right|+\left|2x-4\right|+4\)
e ) \(D=\left|3x-6\right|+\left|3x-9\right|+\left|3x-12\right|+\left|3x-15\right|+2018\)
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
tìm x biết
a) \(\left|2x-7\right|=3\)
b) \(\left|21x-5\right|=\left|3x-7\right|\)
c) \(\left|7-2x\right|=1-3x\)
d) \(\left|x^2+3x\right|+\left|x^2+4x+3\right|=0\)
b)\(\left|21x-5\right|=\left|3x-7\right|\)
\(\Leftrightarrow\begin{cases}21x-5=3x-7\\21x-5=7-3x\end{cases}\)
\(\Leftrightarrow\begin{cases}9x=-1\\24x=12\end{cases}\)
\(\Leftrightarrow\begin{cases}x=-\frac{1}{9}\\x=\frac{1}{2}\end{cases}\)
a)\(\left|2x-7\right|=3\)
\(\Rightarrow2x-7=\pm3\)
Nếu \(2x-7=3\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Nếu \(2x-7=-3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
c)\(\left|7-2x\right|=1-3x\)
\(\Rightarrow7-2x=\pm\left(1-3x\right)\)
Nếu \(7-2x=1-3x\)
\(\Rightarrow x=-6\)
Nếu \(7-2x=-\left(1-3x\right)\)
\(\Rightarrow7-2x=-1+3x\)
\(\Rightarrow5x=8\)
\(\Rightarrow x=\frac{8}{5}\)
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
Tính:
\(a)\left(-2x^2\right)\cdot\left(3x-4x^3+7-x^2\right)\)
\(b)\left(x+3\right)\cdot\left(2x^2-3x-5\right)\)
\(c)\left(-6x^5+7x^4-6x^3\right):3x^3\)
\(d)\left(9x^2-4\right):\left(3x+2\right)\)
\(e)\left(2x^4-13x^3+15x^2+11x-3\right):\left(x^2-4x-3\right)\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
giúp mk với tứ tư mk phải nộp rùi
bài 1:
a, \(2x\left(3x^2-5x+3\right)\)
b, \(-2x\left(x^2+5x-3\right)\)
c, \(\dfrac{-1}{2}x\left(2x^3-4x+3\right)\)
bài 2:
a,\(\left(2x-1\right).\left(x^2-5-4\right)\)
b,\(-\left(5x-4\right).\left(2x+3\right)\)
c,\(\left(2x-y\right).\left(4x^2-2xy+y^2\right)\)
d,\(\left(3x-4\right).\left(x+4\right).\left(5-x\right).\left(2x^2+3x-1\right)\)
e,\(7\left(x-4\right)-\left(7x+3\right).\left(2x^2-x+4\right)\)
bài 3:
c/m rằng gtri của biểu thức ko phụ thuộc vào gtri của biến
a,\(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
b,\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
bài 4 :tìm x biết
a, \(3x+2\left(5-x\right)=0\)
b,\(x\left(2x-1\right).\left(x+5\right)-\left(2x^2+1\right).\left(x+4,5\right)=3,5\)
c,\(3x^2-3x\left(x-2\right)=36\)
d,\(\left(3x^2-x+1\right).\left(x-1\right)+x^2.\left(4-3x\right)=\dfrac{5}{2}\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
Bài 3: (chỉ cần kết quả ko chứa biến là ta có đpcm, nói chung bài này yêu cầu ta rút gọn)
a) \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)
\(=20\)
b) \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
\(=6x-3-5x+15+18x-24-19x\)
\(=-12\)
Đạo hàm của hàm số \(y=\left(-x^2+3x+7\right)^7\) là:
A. \(y'=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
B. \(y'=7\left(-x^2+3x+7\right)^6\)
C. \(y'=\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
D. \(y'=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)