1. Tìm x:a) -5(x-1)+13=2(8-x)-15
b) 3.5x+2+4.5x-3=19.510
c) 6.8x-1+8x+1=821+819.6
2. Rút gọn
(-1/2)3-(3/4)3.(-2)2/ 2.(-1)7+(3/4)2-3/8
3) Cho A=1/4x(x-3). tìm x để
a) A=0
b) A > hoặc= 0
c) A<0
K
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LH
1
20 tháng 10 2021
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
2 tháng 7 2021
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
T
24 tháng 10 2023
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
29 tháng 10 2021
Bài 1:
a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Bài 2:
a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)
\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)
AL
4