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a: (2x-3/2)(|x|-5)=0
=>2x-3/2=0 hoặc |x|-5=0
=>x=3/4 hoặc |x|=5
=>\(x\in\left\{\dfrac{3}{4};5;-5\right\}\)
b: x-8x^4=0
=>x(1-8x^3)=0
=>x=0 hoặc 1-8x^3=0
=>x=1/2 hoặc x=0
c: x^2-(4x+x^2)-5=0
=>x^2-4x-x^2-5=0
=>-4x-5=0
=>x=-5/4
a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)
\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)
=>2x+5=1/44 hoặc 2x+1=-1/44
=>x=-219/88 hoặc x=-221/88
\(Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\)
\(=\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2-\left(3x-3x\right)+\left(1+\dfrac{2}{3}\right)\)
\(=3x^4+2x^2+\dfrac{5}{3}\)
\(3x^4+2x^2+\dfrac{5}{3}=0\)
\(\Rightarrow3x^4+2x^2=-\dfrac{5}{3}\)(Vô lí vì \(3x^4\) và \(2x^2\) luôn lớn hơn hoặc bằng 0)
Vậy Q(x) không có nghiệm
Q(x)=3x^4+2x^2+5/3>=5/3>0 với mọi x
=>Q(x) vô nghiệm
4: \(\Leftrightarrow3^{x+4}\cdot\dfrac{1}{3}-4\cdot3^x=3^{16}\left(1-4\cdot3^3\right)\)
=>\(3^x\cdot27-4\cdot3^x=3^{16}\cdot\left(-107\right)\)
=>3^x*23=3^16*(-107)
=>\(x\in\varnothing\)
2: \(\Leftrightarrow2^x\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\)
=>2^x=2^10
=>x=10
3: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>8^x=8^9
=>x=9
1: \(\Leftrightarrow3^x\cdot\left(4\cdot\dfrac{1}{9}+2\cdot3\right)=3^4\left(4+2\cdot3^3\right)\)
=>3^x=3^4*3^2
=>x=4+2=6
P(x)=-5x^3-1/3+8x^4+x^2
Q(x)=x^4-2x^3+x^2-5x-2/3
P(x)+Q(x)
=x^4-2x^3+x^2-5x-2/3+8x^4-5x^3+x^2-1/3
=9x^4-7x^3+2x^2-5x-1
P(x)-Q(x)
=x^4-2x^3+x^2-5x-2/3-8x^4+5x^3-x^2+1/3
=-7x^4+3x^3-5x-1/3
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
\(A=\dfrac{1-\left|8x-\dfrac{2}{3}\right|}{2}\)
\(\left|8x-\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow1-\left|8x-\dfrac{2}{3}\right|\le1\)
\(MAX_A\Rightarrow MAX_{1-\left|8x-\dfrac{2}{3}\right|}\)
\(MAX_{1-\left|8x-\dfrac{2}{3}\right|}=1\)
Xảy ra khi và chỉ khi:
\(\left|8x-\dfrac{2}{3}\right|=0\Rightarrow8x=\dfrac{2}{3}\Rightarrow x=\dfrac{1}{12}\)
\(\Rightarrow MAX_A=\dfrac{1}{12}\) khi \(x=\dfrac{1}{12}\)
\(B=5-\left|\dfrac{3}{5}-2x\right|+2\)
\(B=7-\left|\dfrac{3}{5}-2x\right|\)
\(\left|\dfrac{3}{5}-2x\right|\ge0\forall x\)
\(\Rightarrow7-\left|\dfrac{3}{5}-2x\right|\le7\)
(ko tìm được MIN đâu nhé)
Dấu "=" xảy ra khi:
\(\left|\dfrac{3}{5}-2x\right|=0\Rightarrow\dfrac{3}{5}=2x\Rightarrow x=\dfrac{3}{10}\)
\(\Rightarrow MAX_B=7\) khi \(x=\dfrac{3}{10}\)
a) ta có : \(\left|8x-\dfrac{2}{3}\right|\ge0\Rightarrow1-\left|8x-\dfrac{2}{3}\right|\le1\Rightarrow\dfrac{1-\left|8x-\dfrac{2}{3}\right|}{2}\le\dfrac{1}{2}\) Vậy GTLN của A=\(\dfrac{1}{2}\) khi và chỉ khi x=\(\dfrac{1}{12}\)
b) Giải tương tự câu a
ĐKXĐ: \(x\ne\pm\dfrac{1}{2}\)
\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
\(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x\left(2x+1\right)}{12\left(2x-1\right)\left(2x+1\right)}-\dfrac{3\left(1+8x\right)\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)
\(\Rightarrow-32x^2=16x^2+8x-3\left(16x^2-6x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-48x^2+18x+3\)
\(\Leftrightarrow-32x^2=-32x^2+26x+3\)
\(\Leftrightarrow26x+3=0\)
\(\Leftrightarrow26x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{26}\) (tmđk)
$Toru$