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a) 2x(x+3) – 3x2(x+2) + x(3x2 + 4x – 6)
= (2x . x + 2x . 3) – (3x2 . x + 3x2 . 2) + (x . 3x2 + x . 4x – x . 6)
= 2x2 + 6x – (3x3 + 6x2) + (3x3 + 4x2 - 6x)
= 2x2 + 6x – 3x3 – 6x2 + 3x3 + 4x2 - 6x
= (– 3x3 + 3x3 ) + (2x2 - 6x2 + 4x2 ) + (6x – 6x)
= 0 + 0 + 0
= 0
b) 3x(2x2 – x) – 2x2(3x+1) + 5(x2 – 1)
= [3x . 2x2 + 3x . (-x)] – (2x2 . 3x + 2x2 . 1) + [5x2 + 5 . (-1)]
= 6x3 – 3x2 – (6x3 +2x2) + 5x2 – 5
= 6x3 – 3x2 – 6x3 - 2x2 + 5x2 – 5
= (6x3 – 6x3 ) + (-3x2 – 2x2 + 5x2) – 5
= 0 + 0 – 5
= - 5
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
a, \(A=2x^2+x+6\)
Với x = 1 suy ra A = 2 + 1 + 6 = 9
Với x = 1/2 suy ra A = 1/2 + 1/2 + 6 = 7
b, \(B=7x-6y-5\)Thay x = 3 ; y = -2 ta được
B = 7.3 - 6 ( - 2 ) - 5 = 21 + 12 - 5 = 33 - 5 = 28
a) cho A(x) = 0
\(=>2x^2-4x=0\)
\(x\left(2-4x\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\4x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)\(B\left(y\right)=4y-8\)
cho B(y) = 0
\(4y-8=0\Rightarrow4y=8\Rightarrow y=2\)
c)\(C\left(t\right)=3t^2-6\)
cho C(t) = 0
\(=>3t^2-6=0=>3t^2=6=>t^2=2\left[{}\begin{matrix}t=\sqrt{2}\\t=-\sqrt{2}\end{matrix}\right.\)
d)\(M\left(x\right)=2x^2+1\)
cho M(x) = 0
\(2x^2+1=0\Rightarrow2x^2=-1\Rightarrow x^2=-\dfrac{1}{2}\left(vl\right)\)
vậy M(x) vô nghiệm
e) cho N(x) = 0
\(2x^2-8=0\)
\(2\left(x^2-4\right)=0\)
\(2\left(x^2+2x-2x-4\right)=0\)
\(2\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
A: Đặt P(x)=0
=>3x-5=0
hay x=5/3
b: Đặt Q(x)=0
=>-2x+6=0
hay x=3
c: Đặt M(y)=0
=>1/2y-3=0
hay y=6
d: Đặt A(x)=0
=>12-3/4x=0
=>3/4x=12
hay x=16
Bài 7
a)cho P(x) = 0
\(=>3x-5=0\Leftrightarrow3x=5\Leftrightarrow x=\dfrac{5}{3}\)
b) cho Q(x) = 0
\(=>6-2x=0\Leftrightarrow2x=6\Leftrightarrow x=3\)
c)cho M(y) = 0
\(=>\dfrac{1}{2}y-3=0\Leftrightarrow\cdot\dfrac{1}{2}y=3\Leftrightarrow y=6\)
d)cho A(x) = 0
\(=>\dfrac{-3}{4}x+12=0=>-\dfrac{3}{4}x=-12=>x=16\)
e)cho B(y) = 0
=>\(2y+15=0=>2y=-15=>y=-\dfrac{15}{2}\)
f) cho C(t) = 0
=>\(2-5t=0=>5t=2=>t=\dfrac{2}{5}\)
\(\hept{\begin{cases}3x=2y\\2x+y=3\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{3}{2}.x\\2x+\frac{3}{2}.x=3\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{3}{2}.x\\\frac{7}{2}.x=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{6}{7}\\y=\frac{9}{7}\end{cases}}}\)
\(\hept{\begin{cases}\frac{x}{3}=\frac{3y}{4}\\3x-y=4\end{cases}\Leftrightarrow\hept{\begin{cases}4x=9y\\3x-y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9y}{4}\\\frac{3.9}{4}y-y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{4}.y\\\frac{23}{4}.y=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{4}.y\\y=\frac{16}{23}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{36}{23}\\y=\frac{16}{23}\end{cases}}}\)
Các phần sau làm tương tự nhé
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
Trả lời:
a, \(\left(3x+y-z\right)-\left(4x-2y+6z\right)=3x+y-z-4x+2y-6z=-x+3y-7z\)
b, \(K=2x\left(-3x+5\right)+3x\left(2x-12\right)+26x=-6x^2+10x+6x^2-36x+26x=0\)
d, \(A=3x^2\left(x-1\right)-\left(3x^2+x\right)=3x^3-3x^2-3x^2-x=3x^3-6x^2+x\)
e, \(B=y\left(2y^2+1\right)-y^2\left(2+2y-y^2\right)=2y^3+y-2y^2-2y^3+y^4=y^4-2y^2+y\)