2^(x+3)+2^x=145-2019^0
giúp tui với help
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\(\left(3x-2\right)^2-6x+4=0\\ =>\left(3x-2\right)^2+2\left(-3x+2\right)=0\\ =>\left(2-3x\right)^2+2\left(2-3x\right)=0\\ =>\left(2-3x\right)\left(2-3x+2\right)=0\\ =>\left(2-3x\right)\left(4-3x\right)=0\\ \)
=> 2-3x=0 hoặc 4-3x=0
Nếu 2-3x=0 thì 3x=2 => \(x=\dfrac{2}{3}\)
Nếu 4-3x=0 thì 3x=4 => \(x=\dfrac{4}{3}\)
Vậy \(x=\dfrac{2}{3},x=\dfrac{4}{3}\)
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Bài 2:
a: Ta có: \(2x^2+y^2-2xy+x+2=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\left(vôlý\right)\)
b: Ta có: \(-x^2-26y^2+10xy-20y-150=0\)
\(\Leftrightarrow x^2-10xy+25y^2+y^2+20y+100+50=0\)
\(\Leftrightarrow\left(x-5y\right)^2+\left(y+10\right)^2+50=0\left(vôlý\right)\)
Bài 1:
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\Leftrightarrow2\left(ab+bc+ca\right)=0-1=-1\)hay \(ab+bc+ca=-\dfrac{1}{2}\Leftrightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}\)Ta có: \(P=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-2.\dfrac{1}{4}=\dfrac{1}{2}\)Vậy \(P=\dfrac{1}{2}\)
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`5/(-x^2+5x-6)+(x+3)/(2-x)=0`
Đk:`x ne 2,x ne 3`
`pt<=>-5/(x^2-5x+6)-(x+3)/(x-2)=0`
`<=>-5-(x+3)(x-3)=0`
`<=>(x+3)(x-3)=-5`
`<=>x^2-9=-5`
`<=>x^2-4=0`
`<=>(x-2)(x+2)=0`
`x ne 2=>x-2 ne 0`
`<=>x+2=0`
`<=>x=-2`
Vậy `S={-2}`
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\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)
=>x=4
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Câu 1. thiếu đề đó bạn ạ
Câu 2:
Ta có: x^3+15x^2+74x+120
=(x^3+4x^2) + (11x^2+44x) + (30x+120)
=(x+4)(x^2+11x+30)
=(x+4)(x+5)(x+6)
Ta có bảng xét dấu
x | -6 | -5 | -4 | ||||
x+4 | - | | | - | | | - | | | + |
x+5 | - | | | - | | | + | | | + |
x+6 | - | | | + | | | + | | | + |
Để (x+4)(x+5)(x+6)<0
Khi có chỉ 1 số âm hoặc cả 3 số âm
<=> x<-6 hoặc -5<x<-4
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\(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow\dfrac{2\sqrt{x}\sqrt{x}}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}=\dfrac{3.2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow\dfrac{2x}{2\sqrt{x}}-\dfrac{6\sqrt{x}}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}=0\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Leftrightarrow...\)
\(\Rightarrow2x+1=6\sqrt{x}\)
\(\Rightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\dfrac{3\pm\sqrt{7}}{2}\)
\(\Rightarrow x=\left(\dfrac{3\pm\sqrt{7}}{2}\right)^2=\dfrac{8\pm3\sqrt{7}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. (2x + 1)2 - 4x2 + 2x2 - 2 = 0
<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x2 - 1) = 0
<=> (4x + 1) + 2x2 - 2 = 0
<=> 4x + 1 + 2x2 - 2 = 0
<=> 2x2 + 4x - 2 + 1 = 0
<=> 2x2 + 4x - 1 = 0
<=> 2x2 + 4x = 1
<=> 2x(x + 2) = 1
Vì 1 chỉ có tích là 1 . 1 nên:
<=> \(\left[{}\begin{matrix}2x=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)
2x+3+2x=145-20190
--> 2x23+2x=145-1
--> 2x(23+1)=144
-->2x=\(\frac{144}{9}\)=16=24
-->x=4