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a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. f(x) = -3x4 + 5x3 + 2x2 - 7x + 7 tại x = 1; 0; 2
xét x=1 có f(x) =-3.14 +5.13 +2.12-7.1+7
=-3.1+5.1+2.1-7+7
=-3+5+2-7+7
=4
xét x=0 có f(x) =-3.04 +5.03 +2.02-7.0+7
=0+0+0-0+7=7
xét x=2 có f(x) =-3.24 +5.23 +2.22-7.2+7
=-3.16+5.8+2.4-14+7
=48+40+8-14+7
=89
2. g(x) = x4 - 5x3 + 7x2 + 15x + 2 tại x = -1; 0; 1; 2
xét x=-1 có: g(x)=(-1)4-5.(-1)3+7.(-1)2+15.(-1)+2
=1-5.(-1)+7.1-15+2
=1-(-5)+7-15+2
=1+5+7-15+2=0
xét x=0 có: g(x)=04-5.03+7.02+15.0+2
=0-0+0+0+2+2=2
xét x=1 có: g(x)=14-5.13+7.12+15.1+2
=1-5.1+7.1-15+2
=1-5+7-15+2
=1-5+7-15+2=-10
xét x=2 có: g(x)=24-5.23+7.22+15.2+2
=32-5.8+7.4-30+2
=32-40+28-30+2
=-8
3. h(x) = -x4 + 3x3 + 2x2 - 5x + 1 tại x = -2; -1; 1; 2
xét x=-2có:h(X)=-(-2)4 + 3(-2)3 + 2.(-2)2 - 5.(-2) + 1
=-(32)+3.(-8)+2.4+10+1
=-32-24+8+10+1
=-37
xét x=2có:h(X)=-(2)4 + 3.23 + 2.22 - 5.2 + 1
=-(32)+3.8+2.4+10+1
=-32+24+8+10+1
=11
xét x=1có:h(X)=14 + 3.13 + 2.12 - 5.1 + 1
=1+3.1+2.1+5+1
=1+3+2+5+1
=13
xét x=-1có:h(X)=-14 + 3.(-1)3 + 2.(-1)2 - 5.(-1) + 1
=1+3.(-1)+2.(-1)+5+1
=1-3-2+5+1
=2
4. r(x) = 3x4 + 7x3 + 4x2 - 2x - 2 tại x = -1; 0; 1
xét x=-1có:r(X)= 3(-1)4 + 7(-1)3 + 4(-1)2 - 2(-1)- 2
= 3.1+7.(-1) +4.1+2-2
=3-7+4+2-2
= 0
xét x=0có:r(X)= 3.04 + 7.03 + 4.02 - 2.0- 2
= 0+0+0-0-2
= -2
xét x=1có:r(X)= 3(1)4 + 7(1)3 + 4(1)2 - 2(1)- 2
= 3.1+7.1 +4.1-2-2
=3+7+4-2-2
= 10
![](https://rs.olm.vn/images/avt/0.png?1311)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^5-3*x^2-(7*x^4-9*x^3+x^2-1/4*x+5*x^4-x^5+x^2-2*x^3+3*x^2-1/4)=0
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
Sửa đề: tại x=-1/2
Đặt A=-5x+7x-3x-2x
=x(-5+7-3-2)
=-3x
Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=-3x\cdot\dfrac{-1}{2}=\dfrac{3}{2}\)
b: Sửa đề: \(-4x^2-3x^2+2x^2-x^2\); tại x=-1/2
Đặt \(B=-4x^2-3x^2+2x^2-x^2\)
\(=x^2\left(-4-3+2-1\right)=-6x^2\)
Thay x=-1/2 vào B, ta được:
\(B=-6\cdot\left(-\dfrac{1}{2}\right)^2=-6\cdot\dfrac{1}{4}=-\dfrac{6}{4}=-\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|3x-2\right|=x+1\left(1\right)\)
\(ĐK:x+1\ge0\Rightarrow x\ge-1\)
Với \(x\ge-1\) thì \(\left(1\right)\Rightarrow\left[{}\begin{matrix}3x-2=x+1\\3x-2=-\left(x+1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=1+2\\3x-2=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\3x+x=-1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\4x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=0,25\end{matrix}\right.\)
Vậy \(x\in\left\{1,5;0,25\right\}\)
\(\left|2x-1\right|-3+5x=7x-2\)
\(\Rightarrow\left|2x-1\right|=7x-2+3-5x\)
\(\Rightarrow\left|2x-1\right|=\left(7x-5x\right)-2+3\)
\(\Rightarrow\left|2x-1\right|=2x+1\left(2\right)\)
\(ĐK:2x+1\ge0\Rightarrow2x\ge-1\Rightarrow x\ge\dfrac{-1}{2}\)
Với \(x\ge\dfrac{-1}{2}\) thì
\(\left(2\right)\Rightarrow\left[{}\begin{matrix}2x-1=2x+1\\2x-1=-\left(2x+1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-2x=1+1\\2x-1=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2(loại)\\2x+2x=-1+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(loại\right)\\4x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}0=2\left(loại\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy \(x=0\)