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a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
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
a: =>x=(-2/3)^5:(-2/3)^2=(-2/3)^3=-8/27
b: =>x*(-1/3)^3=(-1/3)^4
=>x=-1/3
d: =>3x-2=-3
=>3x=-1
=>x=-1/3
a: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
A=(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=(x2+5x+4) ta có:
t(t+2)-24=t2+6t-2t-24
=t(t+6)-4(t+6)
=(t-4)(t+6).Thay vào ta đc:
(x2+5x+4-4)(x2+5x+4+6)=(x2+5x)(x2+5x+10)
=x(x+5)(x2+5x+10)
B=(x2+3x+2)(x2+7x+120-24)
=(x2+3x+2)(x2+7x+96)
=(x2+2x+x+2)(x2+7x+96)
=[x(x+2)+(x+2)](x2+7x+96)
=(x+1)(x+2)(x2+7x+96)
C và D bn cx lm tương tự
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 dễ tự làm:
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^n.3^3+3^n.3^1+2^n.2^3+2^n.2^2\)
\(=3^n.9+3^n.3+2^n.8+2^n.4\)
\(=3^n\left(9+3\right)+2^n\left(8+4\right)\)
\(=3^n.12+2^n.12\)
\(=12\left(3^n+2^n\right)\)
\(=6.2\left(3^n+2^n\right)⋮6\rightarrowđpcm\)
https://hoc24.vn/hoi-dap/question/462129.html?pos=1207827
vào link này làm hộ mik bài 3 nha
\(a\)\(,\)\(\left(2x-3\right)^2\)\(=\)\(4^2\)(1)
mà ta có \(4^2\)=\(\left(-4\right)^2\)(2)
Từ (1) và (2)\(\Rightarrow\)\(\left(2x-3\right)^2\)=\(4^2\)=\(\left(-4\right)^2\)
\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-1}{2}\end{cases}}\)(thỏa mãn \(x\)\(\in\)\(Q\))
Vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-1}{2}\end{cases}}\)
\(b,\)\(\left(3x-2\right)^5\)\(=\)\(-243\)
\(\Rightarrow\)\(\left(3x-2\right)^5\)\(=\)\(\left(-3\right)^5\)
\(\Rightarrow\)\(3x-2=-3\)
\(\Rightarrow\)\(3x=-1\)
\(\Rightarrow\)\(x=\frac{-1}{3}\)(thỏa mãn \(x\in Q\))
Vậy \(x=\frac{-1}{3}\)
\(c,\)\(\left(7x+2\right)^{-1}=3^{-2}\)
\(\Rightarrow\frac{1}{7x+2}=\frac{1}{3^2}\)
\(\Rightarrow\frac{1}{7x+2}=\frac{1}{9}\)
\(\Rightarrow\)\(7x+2=9\)
\(\Rightarrow\)\(7x=7\)
\(\Rightarrow x=1\)(thỏa mãn \(x\in Q\))
Vậy \(x=1\)
A,\(\left(2x-3\right)^2=4^2\)
\(2x-3=4\)
\(2x=7\)
\(x=3,5\)
Tương tự