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F(x)=62+5x+8+3x-3x2+3x3
=(36+8)+(5x+3x)-3x2+3x3
=3x3-3x2+8x+44
G(x)=12x2-6-9x2+3x3
=3x3+(12x2-9x2)-6
=3x3+3x2-6
F(x)+G(x)=3x3-3x2+8x+44+3x3+3x2-6
=(3x3+3x3)+(-3x2+3x2)+8x+(44-6)
=6x3+8x+38
\(F\left(x\right)=G\left(x\right)\\ \Rightarrow6^2-5x+8+3x-3x^2+3x^3=12x^2-6-9x^2+3x^3\\ \Leftrightarrow-3x^2-2x+44=3x^2-6\\ \Leftrightarrow6x^2+2x-50=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{301}}{6}\\x=\dfrac{-1-\sqrt{301}}{6}\end{matrix}\right.\)
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Đặt f(x)=0
nên 3x-6=0
hay x=2
Đặt h(x)=0
nên 30-5x=0
hay x=6
Đặt g(x)=0
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Ta có:
\(f\left(x\right)=0\Leftrightarrow3x-6=0\Leftrightarrow x=2\)
Vậy nghiệm của đa thức f(x) là 2
\(h\left(x\right)=0\Leftrightarrow-5x+30=0\Leftrightarrow x=6\)
Vậy nghiệm của đa thức h(x) là 6
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(\Leftrightarrow\left|3x-7\right|+\left|3x-15\right|=8\)
TH1: x<7/3
Pt sẽ là \(7-3x+15-3x=8\)
=>22-6x=8
=>6x=14
hay x=7/3(loại)
TH2: 7/3<=x<5
Pt sẽ là \(3x-7+15-3x=8\)
=>8=8(luôn đúng)
TH3: x>=5
Pt sẽ là 3x-7+3x-15=8
=>6x-22=8
hay x=5(nhận)
b: \(\Leftrightarrow\left|4x-98\right|+\left|4x-8\right|=90\)
TH1: x<2
Pt sẽ là 8-4x+98-4x=90
=>106-8x=90
=>x=2(loại)
TH2: 2<=x<49/2
Pt sẽ là 4x-8+98-4x=90
=>90=90(luôn đúng)
TH3: x>=49/2
Pt sẽ là 4x-8+4x-98=90
=>8x-106=90
=>8x=196
hay x=24,5(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a, \(F\left(x\right)=-3x+6\)
\(F\left(x\right)=-3x+6=0\)
\(\Rightarrow x=2\)
Vậy, .....
b, \(G\left(x\right)=5x-10\)
\(G\left(x\right)=5x-10=0\)
\(\Rightarrow x=2\)
Vậy, ....
d, \(M\left(x\right)=x^2+7x-8\)
\(M\left(x\right)=x^2+7x-8=0\)
\(\Rightarrow\left(x^2+8x\right)-\left(x+8\right)=0\)
\(\Rightarrow x\left(x+8\right)-\left(x+8\right)=0\)
\(\Rightarrow\left(x+8\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)
Vậy, ....
e, \(L\left(x\right)=5x^2+9x+4\)
\(L\left(x\right)=5x^2+9x+4=0\)
\(\Rightarrow\left(5x^2+5x\right)+\left(4x+4\right)=0\)
\(\Rightarrow5x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(4+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\4+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\dfrac{-4}{5}\end{matrix}\right.\)
Vậy, ....
F(x)=-3x+6=0
<=>-3(x-2)=0
<=>x-2=0
<=>x=2
Vậy...
G(x)=5x-10=0
<=>5(x-2)=0
<=>x-2=0
<=>x=2
Vậy...
K(sai đề)
M(x)=x2+7x-8=0
<=>(x2+8x)-(x+8)=0
<=>x(x+8)-(x+8)=0
<=>(x+8)(x-1)=0
<=>x+8=0 hoặc x-1=0
<=>x=-8 hoặc x=1
Vậy...
L(x)=5x2+9x+4=0
<=>(5x2+5x)+(4x+4)=0
<=>5x(x+1)+4(x+1)=0
<=>(x+1)(5x+4)=0
<=>x+1=0 hoặc 5x+4=0
<=>x=-1 hoặc x=\(-\dfrac{4}{5}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0
\(\left(5x-6\right)^2-\left(3x-7\right)^2=-9x\left(x-8\right)+\left(5x-6\right)^2-13\)
\(\Rightarrow25x^2-60x+36-9x^2+42x-49=-9x^2+72x+25x^2-60x+36-13\)
\(\Rightarrow\left(25x^2-25x^2\right)-\left(60x-60x\right)+\left(36-36\right)-\left(9x^2-9x^2\right)+\left(42x-72x\right)-\left(49-13\right)=0\)
\(\Rightarrow-30x-36=0\)
\(\Rightarrow-30x=36\)
\(\Rightarrow x=-\dfrac{36}{30}\)
\(\Rightarrow x=-\dfrac{6}{5}\)
Vậy: \(x=-\dfrac{6}{5}\)