\(\left(2x+7\right)^2=x-12\)
Tìm x
\(2x.\left(x-5\right)-x.\left(3+2x\right)=26\)
\(\left(x-7\right).\left(x-5\right)-12.\left(3x-7\right)=15\)
\(4.\left(18-5x\right)-12.\left(3x-7\right)=15.\left(2x-16\right)-6.\left(x+14\right)\)
\(\left(x-1\right).\left(x^2+x+1\right)=x^3-2x\)
tìm x biết : x.(x+2/3)-x.(x-3/4)=7/12
b)\(\sqrt{x^2}+1\)=x+2
c)\(\left(2x+1\right)^5\)=\(\left(2x+1\right)^{2019}\)
mọi người ơi câu b là giá trị tuyệt đối của x^2 -1 nha
giúp mình mình tick cho
a) \(\Leftrightarrow x^2+\dfrac{2}{3}x-x^2+\dfrac{3}{4}x=\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{17}{12}x=\dfrac{7}{12}\Leftrightarrow x=\dfrac{7}{17}\)
c) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=-1\\2x+1=1\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Giải phương trình
\(\frac{13}{\left(2x+7\right)\left(x-3\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
\(\left(1-\frac{2x-1}{x+1}\right)+6\left(1-\frac{2x-1}{x+1}\right)^2=\frac{12\left(2x-1\right)}{x+1}-20\)
13(x+3)+(x+3)(x-3)=6(2x+7)
13x+39+x^2-9-12x-42=0
x^2+x-12=0
x=3 và x=-4
**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Giải các phương trình sau :
a) \(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\)
b) \(\left(1-\dfrac{2x-1}{x+1}\right)^3+6\left(1-\dfrac{2x-1}{x+1}\right)^2=\dfrac{12\left(2x-1\right)}{x+1}-20\)
GPT: \(\frac{x^2-2x+14}{\sqrt{\left(7-2x\right)\left(2x+3\right)}}+\frac{12+2x-x^2}{\sqrt{4x^2-8x+29}}=20\)
Tìm số nguyên x
1/ \(17x+3\left(-16x-37\right)=2x+43-4x\)
2/ \(-2x-3\left(x-17\right)=34-2\left(-x+25\right)\)
3/ \(\left\{-3x+2\left[45-x-3\left(3x+7\right)-2x\right]+4x\right\}=55-103-57:\left[-2\left(2x-1\right)^2-\left(-9\right)^0\right]=-106\)
4/ \(-2x+3\left\{12-2\left[3x-\left(20+2x\right)-4x\right]+1\right\}=45\)
5/ \(3x-32>-5+1\)
6/ \(15+4x< 2x-145\)
7/ \(-3\left(2x+5\right)-16< -4\left(3-2x\right)\)
8/ \(-2x+15< 3x-7< 19-x\)
bài tập tết nâng cao phải ko
mk cũng có nhưng chưa làm dc
a) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{3}-\left(x-1\right)\)
b) \(x^2-6x-2+\dfrac{14}{x^2-6x+7}=0\)
c) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
d) \(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}=\dfrac{6}{x^2-9}\)
e) \(\left(1-\dfrac{2x-1}{x+1}\right)^3+6\left(1-\dfrac{2x-1}{x+1}\right)^2=\dfrac{12\left(2x-1\right)}{x+1}-20\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
4,\(\dfrac{x+1}{3}\)+\(\dfrac{3\left(2x+1\right)}{4}\)=\(\dfrac{2x+3\left(x+1\right)}{6}\)+\(\dfrac{7+12x}{12}\)
5,\(\dfrac{2x}{3}\)+\(\dfrac{2x-1}{6}\)=4-\(\dfrac{x}{3}\)
6,\(\dfrac{x-1}{2}\)+\(\dfrac{x-1}{4}\)=1-\(\dfrac{2\left(x-1\right)}{3}\)
4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)
\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm
5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)
6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)
\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)
Câu 1: Tìm x:
a. \(\left(x-1\right)\left(x+2\right)-x^2+3=5\)
b. \(\left(2x+1\right)\left(x-3\right)-2x\left(x+7\right)=100\)
c. \(\left(3x-1\right)\left(x+2\right)-\left(2-3x\right)\left(x+3\right)=12\)
Câu 2: Chứng minh giá trị biểu thức sau không phụ thuộc vào biến
\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)
Câu 1:
a. \(\left(x-1\right)\left(x+2\right)-x^2+3=5\)
\(x^2+2x-x-2-x^2+3=5\)
\(x+1=5\)
\(x=4\)
b. \(\left(2x+1\right)\left(x-3\right)-2x\left(x+7\right)=100\)
\(2x^2-6x+x-3-2x^2-14x=100\)
\(-19x-3=100\)
\(x=\frac{103}{-19}\)
\(x=-7\)
c. \(\left(3x-1\right)\left(x+2\right)-\left(2-3x\right)\left(x+3\right)=12\)
\(3x^2+6x-x-2-\left(2x+6-3x^2-9x\right)=12\)
\(3x^2+6x-x-2-2x-6+3x^2+9x=12\)
\(6x^2+12x-8=12\)
\(6x^2+12x=20\)
Câu 2:
\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)
\(=2x^2+3x-10x-15-2x^2+6x+x+7\)
\(=-8\) (không phụ thuộc vào biến)
xét tính đồng biến nghịch biến
a) \(y=\sqrt{x^2-4x-3}\)
b) \(y=\sqrt{x^3-4x^2}\)
c) \(y=\left(2x+3\right)^{12}\left(6-5x\right)^9\left(x-7\right)^5\)
d) \(y=\sqrt{2x^3-3x^2}\)