Giải hệ phương trình :
x3 + y -\(\dfrac{1}{z}\) = 1
\(2x^3-\sqrt{3}y-\dfrac{\sqrt{2}}{z}=\sqrt{2}\)
\(3x^3-\sqrt{5}y+\dfrac{2}{z}=3\)
Giải trên máy casio
giải các phương trình sau:
a. \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\)
b. \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c. \(\sqrt{\dfrac{3x-2}{x+1}}=3\)
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
1) \(\dfrac{x-3x^2}{2}+\sqrt{2x^4-x^3+7x^2-3x+3}=2\)
2) \(1+\sqrt{\dfrac{x-2}{1-x}}=\dfrac{2x^2-2x+1}{x^2-2x+2}\)
3) \(x+y+z+\dfrac{3}{x-1}+\dfrac{3}{y-1}+\dfrac{3}{z-1}=2\left(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}\right)\) với x ,y ,z > 1
4) \(\sqrt[3]{x+6}+x^2=7-\sqrt{x-1}\)
5) \(x^4-2x^3+x-\sqrt{2\left(x^2-x\right)}=0\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\left(xy-2\right)^2+6y=3\left(\dfrac{1}{x}-\dfrac{3}{x^2}\right)\\y^3-4y^2+\dfrac{6}{x}+\left(y-1\right)\sqrt{\left(3y-2\right)}=\dfrac{9}{x^2}\end{matrix}\right.\)
Giải hệ phương trình: \(\begin{cases}3\sqrt[3]{3x^2+y+1}=\left(x-1\right)^3-y\\x^3-y-2x^2+2x+\sqrt{x}=\sqrt{x^3-y-2x^2+2x+21}\end{cases}\)
Cho x,y,z dương thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\) . Chứng minh rằng \(\dfrac{1}{\sqrt{2x^2+y^2+3}}+\dfrac{1}{\sqrt{2y^2+z^2+3}}+\dfrac{1}{\sqrt{2z^2+x^2+3}}\) ≤ \(\dfrac{\sqrt{6}}{2}\)
\(VT^2\le3\left(\dfrac{1}{2x^2+y^2+3}+\dfrac{1}{2y^2+z^2+3}+\dfrac{1}{2z^2+x^2+3}\right)\)
Mặt khác:
\(\dfrac{1}{2\left(x^2+1\right)+y^2+1}\le\dfrac{1}{4x+2y}=\dfrac{1}{2}\left(\dfrac{1}{x+x+y}\right)\le\dfrac{1}{18}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow VT^2\le\dfrac{1}{6}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\)
\(\Rightarrow VT\le\dfrac{\sqrt{6}}{2}\)
Giải hệ phương trình: \(\begin{cases}\sqrt{2y^2+3x+1}+\sqrt{1-3x}=2\sqrt{y^2+1}\\y^3+1+\sqrt[3]{y^3-3x^2+3x-1}=x\left(y^2+1\right)\end{cases}\)
Giải hệ bất phuơng trình:
\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\dfrac{y}{x}=\dfrac{2\sqrt{x}}{y}+2\\y\left(\sqrt{x^2+1}-1\right)=\sqrt{3x^2+3}\end{matrix}\right.\)
ĐKXD: x, y > 0.
\(Pt_{\left(1\right)}\Leftrightarrow\dfrac{y+\sqrt{x}}{x}=\dfrac{2\left(y+\sqrt{x}\right)}{y}\Leftrightarrow\left(y+\sqrt{x}\right)\left(\dfrac{1}{x}-\dfrac{2}{y}\right)=0\)
\(\Rightarrow y=2x\), thế vào Pt(2): \(2x\left(\sqrt{x^2+1}-1\right)=\sqrt{3x^2+3}\)
Đặt \(\sqrt{x^2+1}=a\) thì \(\left\{{}\begin{matrix}2x\left(a-1\right)=\sqrt{3}a\\a^2-x^2=1\end{matrix}\right.\)
Giải ra ta được \(\left(a-2\right)\left(4a^3-3a+2\right)=0\) nhưng vì \(a\ge1\) nên a=2
Do đó \(x=\pm\sqrt{3}\). Vậy (x, y)=...
Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
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x5=y7=z3=x225=y249=z29" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27