4x-3y=6

3y+4x=10

3y+4x-4x+3y=10-6

6y=4

y=2/3

thay vào 4x-3.2/3=6

4x-2=6

4x=8

x=2

\(\hept{\begin{cases}4x-3y=6\\4x+3y=10\end{cases}\Leftrightarrow\hept{\begin{cases}-6y=-4\\4x+3y=10\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{2}{3}\\4x+3y=10\end{cases}}}\)

Thay y = 2/3 vào pt 2 ta được : \(4x+2=10\Leftrightarrow x=2\)

Vậy hpt có một nghiệm là ( x ; y ) = ( \(2;\frac{2}{3}\))

a) đk: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

Ta có: 

\(P=\frac{x-2\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\)

\(P=\frac{x\sqrt{x}-2x+x-1+1+2x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\sqrt{x}}\)

\(P=\frac{x\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\sqrt{x}}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\sqrt{x}}=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

b) Đặt \(\sqrt{x}=a>0\)

\(\Rightarrow P=\frac{a+2}{a^2+a+1}\Leftrightarrow a^2P+aP+P=a+2\)

\(\Leftrightarrow a^2P+a\left(P-1\right)+\left(P-2\right)=0\)

\(\Delta=\left(P-1\right)^2-4P\left(P-2\right)=-3P^2+6P+1\)

\(\Rightarrow\Delta\ge0\Leftrightarrow-3P^2+6P+1\ge0\Leftrightarrow\frac{3+2\sqrt{3}}{3}\ge P\ge\frac{3-2\sqrt{3}}{3}\)

\(\Leftrightarrow2\ge P\ge0\)

Nếu \(P=0\Leftrightarrow\frac{\sqrt{x}+2}{x+\sqrt{x}+1}=0\Rightarrow\sqrt{x}=-2\left(ktm\right)\)

Nếu \(P=1\Leftrightarrow\sqrt{x}+2=x+\sqrt{x}+1\Leftrightarrow x=1\)(ktm)

Nếu \(P=2\Leftrightarrow\sqrt{x}+2=2x+2\sqrt{x}+2\Leftrightarrow2x+\sqrt{x}=0\)

\(\Leftrightarrow\left(2\sqrt{x}+1\right)\sqrt{x}=0\Rightarrow x=0\left(ktm\right)\)

Vậy không tồn tại giá trị của x để P nguyên

a, Với \(x\ge0;x\ne1\)

\(P=\frac{x-2\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\frac{1+2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{x\sqrt{x}-2\sqrt{x}+x-1+1+2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x\sqrt{x}+3x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{x+3\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}+4}{x+\sqrt{x}+1}\)

1.236067977

1.2360679777

\(\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)do \(\sqrt{5}-1>0\)

\(\frac{2+\sqrt{2}}{1+\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}+2-\sqrt{2}\)do \(2-\sqrt{2}>0\)

\(=\sqrt{2}+2-\sqrt{2}=2\)

a2+b2+c2=4−abc≤4a2+b2+c2=4−abc≤4

Smax=4Smax=4 khi 1 trong 3 số bằng 0

4=abc+a2+b2+c2≥abc+33√(abc)24=abc+a2+b2+c2≥abc+3(abc)23

Đặt 3√abc=x>0⇒x3+3x2−4≤0abc3=x>0⇒x3+3x2−4≤0

⇔(x−1)(x+2)2≤0⇒x≤1⇔(x−1)(x+2)2≤0⇒x≤1

⇒abc≤1⇒S=4−abc≥3⇒abc≤1⇒S=4−abc≥3

Dấu "=" xảy ra khi a=b=c=1

a: Xét tứ giác ODAE có

góc ODA+góc OEA=180 độ

=>ODAE là tứ giác nội tiếp

b: \(AE=\sqrt{\left(3R\right)^2-R^2}=2\sqrt{2}\cdot R\)

\(OI=\dfrac{OE^2}{OA}=\dfrac{R^2}{3R}=\dfrac{R}{3}\)

c: Xét ΔDIK vuông tại I và ΔDHE vuông tại H có

góc IDK chung

=>ΔDIK đồng dạng vơi ΔDHE

=>DI/DH=DK/DE

=>DH*DK=DI*DE=2*IE^2

Ta có: 

\(a=\sqrt[3]{7+\sqrt{50}}=\sqrt[3]{7+5\sqrt{2}}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}=1+\sqrt{2}\)

\(b=\sqrt[3]{7-\sqrt{50}}=\sqrt[3]{7-5\sqrt{2}}=\sqrt[3]{\left(1-\sqrt{2}\right)^3}=1-\sqrt{2}\)

\(\Rightarrow M=a+b=2\) là số chẵn (đpcm)

Lại có:

\(a+b=2;a.b=\left(1+\sqrt{2}\right).\left(1-\sqrt{2}\right)=-1\)\(;a^2+b^2=\left(a+b\right)^2-2ab=6\)

\(N=a^7+b^7\)

\(=\left(a^7+a^4b^3\right)+\left(b^7+a^3b^4\right)-\left(a^4b^3+a^3b^4\right)\)

\(=a^4\left(a^3+b^3\right)+b^4\left(a^3+b^3\right)-a^3b^3\left(a+b\right)\)

\(=\left(a^3+b^3\right)\left(a^4+b^4\right)+2\)

\(=\left(a+b\right)\left(a^2+b^2-ab\right)\left[\left(a^2+b^2\right)^2-2a^2b^2\right]+2\)

\(=2\left(7.34+1\right)=478\) là số chẵn(đpcm)