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Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
a. \(2.\left(5x-8\right)-3.\left(4x-5\right)=4.\left(3x-4\right)+11\Leftrightarrow10x-16-12x+15=12x-16+11\\ \)
\(\Leftrightarrow-2x-1=12x-5\Leftrightarrow14x-4=0\Leftrightarrow x=\frac{2}{7}\)
\(a,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=-16+11+16-15\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\frac{-4}{-14}=\frac{2}{7}\)
A=x2-4x+7
= x2-4x+4+3
= (x-2)2+3
Vì (x+2)2>/ 0
Nên (x-2)2+3>/3
Vậy MAX của A=3 khi x-2=0 => x=2
2)
a) \(x^3-5x^2+8x-4=0\)
\(\Leftrightarrow x^3-4x^2-x^2+4x+4x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy x=1 ; x=2
b) \(2x^3-x^2+3x+6=0\)
\(\Leftrightarrow2x^3-2x-x^2-x+6x+6=0\)
\(\Leftrightarrow\left(2x^3-2x\right)-\left(x^2+x\right)+\left(6x+6\right)=0\)
\(\Leftrightarrow2x\left(x^2-1\right)-x\left(x+1\right)+6\left(x+1\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)-x\left(x+1\right)+6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x-x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2-3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x^2-3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x=-6\left(loai\right)\end{matrix}\right.\)
Vậy x=-1
a/ \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
<=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
<=> \(\left(2x+3\right)^2-4x^2+1=22\)
<=> \(\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
<=> \(3\left(4x+3\right)=21\)
<=> \(4x+3=7\)
<=> \(4x=4\)
<=> \(x=1\)
......................?
mik ko biết
mong bn thông cảm
nha ................
a: \(\Leftrightarrow2x^2-6x+x-3=5x+4-2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2-5x-3=5x+4-2x^2+8\)
\(\Leftrightarrow4x^2-10x-9=0\)
\(\text{Δ}=\left(-10\right)^2-4\cdot4\cdot\left(-9\right)=100+144=244>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{10-\sqrt{244}}{8}=\dfrac{5-\sqrt{61}}{4}\\x_2=\dfrac{5+\sqrt{61}}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow3x-4x^2-4x-1=4x^2-\left(2x^2+2x-x-1\right)\)
\(\Leftrightarrow-4x^2-x-1-4x^2+2x^2+x-1=0\)
\(\Leftrightarrow-6x^2-2=0\)
hay \(x\in\varnothing\)
c: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{x^2+2x-3}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=x^2+2x-3-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+2x-7\)
=>5x+2=2x-7
=>3x=-9
hay x=-3(loại)
\(a,\left|15+x\right|+x=-15\)
\(\Rightarrow\left|15+x\right|=-15-x\)
\(\Rightarrow\left|15+x\right|=-\left(15+x\right)\)
Vì \(\left|15+x\right|\ge0\forall x;-\left(15+x\right)\le0\forall x\)
\(\Rightarrow15+x=-15-x=0\Rightarrow x=-15\)