Tìm x:
\(\left(4x-1\right)^2=\left(1-4x\right)^4\)
Những câu hỏi liên quan
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Tìm Min B = \(\sqrt{4x^4-4x^2\left(x+1\right)+\left(x+1\right)^2+9}\).
\(B=\sqrt{4x^4-4x^2\left(x+1\right)+\left(x+1\right)^2+9}\)
\(=\sqrt{\left(2x^2-x-1\right)^2+9}\)\(\ge\sqrt{9}=3\)
Dấu "=" xảy ra khi \(2x^2-x-1=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(B_{min}=3\)
Đúng 1
Bình luận (0)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Tìm x biết:
a) \(3x^2-4x=0\). b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\).
c) \(9x^2+6x+1=0\). d) \(x^2-4x=4\).
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
Đúng 0
Bình luận (0)
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Tìm x, y, zϵ R biết: \(\left(4x^2-4x+1\right)^{2022}+\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)
mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Tìm giá trị nhỏ nhất của các biểu thức sauAx^2-4x+1 B4x^2+4x+11 Cleft(x-1right)left(x+3right)left(x+2right)left(x+6right)D2x^2+y^2-2xy+2x-4y+9 Tìm giá trị lớn nhất của các biểu thức sauE5-8x-x^2F4x-x^2+1
Đọc tiếp
Tìm giá trị nhỏ nhất của các biểu thức sau
A=\(x^2-4x+1\) \(B=4x^2+4x+11\)
\(C=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)
\(D=2x^2+y^2-2xy+2x-4y+9\)
Tìm giá trị lớn nhất của các biểu thức sau
\(E=5-8x-x^2\)
\(F=4x-x^2+1\)
\(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\)
\(\dfrac{1}{x+2}+\dfrac{3}{x^2-4}+\dfrac{x-14}{\left(x^2+4x+4\right).\left(x-2\right)}\)
\(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\dfrac{1}{x+3}+\dfrac{1}{\left(x+3\right).\left(x+2\right)}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
Đúng 1
Bình luận (0)
Tìm GTNN của:B=\(\sqrt{4x^4-4x^2\left(x+1\right)+\left(x+1\right)^2+9}\)
\(\sqrt{\left(2x^2-x-1\right)^2+9}\ge\sqrt{9}=3\)
min B =3 \(\Leftrightarrow2x^2-x-1=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}\)
Đúng 0
Bình luận (0)
27 tháng 3 2022 lúc 20:25
Giải các phương trình:
a) (\(x^2-4x\))\(^2\) = 4\(\left(x^2-4x\right)\)
b) \(\left(x+2\right)^2-x+1=\left(x-1\right)\left(x+1\right)\)
a) (x2 - 4x)2 = 4(x2 - 4x)
<=> (x2 - 4x)(x2 - 4x - 4) = 0
<=> x(x - 4)(x2 - 4x - 4) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\\left(x-2\right)^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=\pm\sqrt{8}+2\end{matrix}\right.\)
b) (x + 2)2 - x + 1 = (x - 1)(x + 1)
<=> x2 + 4x + 4 - x + 1 = x2 - 1
<=> 3x + 5 = -1
<=> x = -2
Đúng 1
Bình luận (0)
Phân tích đa thức sau thành nhân tử
\(6x^2\left(x-4\right)^2-60\left(x^2-4x-1\right)+90\)
Tìm x, biết
\(4\left(4x-5\right)^2-16x^2+25=\left(5-4x\right)\left(2x-3\right)\)