Chương 2: HÀM SỐ LŨY THỪA. HÀM SỐ MŨ VÀ HÀM SỐ LÔGARIT - Hỏi đáp

1.

ĐKXĐ: \(x< 1\)

\(\Leftrightarrow\log_2\left(1-x\right)\left(3-x\right)=3\)

\(\Leftrightarrow\left(1-x\right)\left(3-x\right)=8\)

\(\Leftrightarrow x^2-4x-5=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=5\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{-1\right\}\)

2.

\(\log_41458=\log_{2^2}\left(2.3^6\right)=\frac{1}{2}\log_2\left(2.3^6\right)\)

\(=\frac{1}{2}\log_22+\frac{1}{2}\log_23^6=\frac{1}{2}+3\log_23\)

\(=\frac{1}{2}+3a\)

\(\Leftrightarrow\left(3^x\right)^2+\left(3x-7\right)3^x-3x+6=0\)

Đặt \(3^x=t>0\)

\(\Rightarrow t^2+\left(3x-7\right)t-3x+6=0\)

\(\Leftrightarrow t^2-1+\left(3x-7\right)t-\left(3x-7\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)+\left(3x-7\right)\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1+3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t+3x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3^x=1\\3^x+3x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3^x+3x-6=0\left(1\right)\end{matrix}\right.\)

Xét (1), nhận thấy \(x=1\) là 1 nghiệm

Xét hàm \(f\left(x\right)=3^x+3x-6\Rightarrow f'\left(x\right)=3^xln3+3>0\) ;\(\forall x\)

\(\Rightarrow f\left(x\right)\) đồng biến trên R \(\Rightarrow f\left(x\right)=0\) có tối đa 1 nghiệm

\(\Rightarrow\left(1\right)\) có nghiệm duy nhất \(x=1\)

Vậy \(x=\left\{0;1\right\}\)

\(5x^2+10yz\le5\left(x^2+y^2+z^2\right)=9x\left(y+z\right)+18yz\)

\(\Leftrightarrow5x^2\le9x\left(y+z\right)+8yz\le9x\left(y+z\right)+2\left(y+z\right)^2\)

\(\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-9\left(\frac{x}{y+z}\right)-2\le0\)

\(\Leftrightarrow\left(\frac{5x}{y+z}+1\right)\left(\frac{x}{y+z}-2\right)\le0\)

\(\Leftrightarrow\frac{x}{y+z}\le2\Rightarrow x\le2y+2z\)

\(\Rightarrow x+y+z\le3\left(y+z\right)\)

\(P\le\frac{2x}{\left(y+z\right)^2}-\frac{1}{\left(x+y+z\right)^3}\le\frac{4\left(y+z\right)}{\left(y+z\right)^2}-\frac{1}{\left(3y+3z\right)^3}\)

\(P\le\frac{4}{\left(y+z\right)}-\frac{1}{27\left(y+z\right)^3}=\frac{4}{y+z}-\frac{1}{27\left(y+z\right)^3}-16+16\)

\(P\le-\frac{1}{27}\left(\frac{1}{y+z}-6\right)^2\left(\frac{1}{y+z}+12\right)+16\le16\)

\(P_{max}=16\) khi \(\left(x;y;z\right)=\left(\frac{1}{3};\frac{1}{12};\frac{1}{12}\right)\)

1.

ĐKXĐ: \(x>\frac{1}{2}\)

\(log_3\left(2x-1\right)=-2\Leftrightarrow2x-1=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-8\Rightarrow x=-\frac{7}{2}< \frac{1}{2}\left(ktm\right)\)

Vậy pt vô nghiệm

2. ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)

\(log\left(x^2-x-2\right)=1\Leftrightarrow x^2-x-2=10\)

\(\Leftrightarrow x^2-x-12=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

\(\left(x^2+y^2+1\right)^2+1\le\left(x^2+y^2+1\right)^2+3x^2y^2+1=4x^2+5y^2\le5x^2+5y^2\)

\(\Rightarrow\left(x^2+y^2+1\right)^2-5\left(x^2+y^2+1\right)+6\le0\)

\(\Leftrightarrow2\le x^2+y^2+1\le3\)

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

Đặt \(x^2+y^2+1=t\Rightarrow t\in\left[2;3\right]\)

\(\Rightarrow P=\frac{t^2+4}{t}-3=t+\frac{4}{t}-3\ge2\sqrt{\frac{4t}{t}}-3=1\)

\(P_{min}=1\) khi \(\left(x^2;y^2\right)=\left(0;1\right)\)

\(P=\frac{t^2+4}{t}-\frac{13}{3}+\frac{4}{3}=\frac{3t^2-13t+12}{3t}+\frac{4}{3}=\frac{\left(t-3\right)\left(3t-4\right)}{3t}+\frac{4}{3}\le\frac{4}{3}\)

\(P_{max}=\frac{4}{3}\) khi \(\left(x^2;y^2\right)=\left(0;2\right)\)

đk:x >0

Pt\(\Leftrightarrow\) X³ +2X =3

\(\Leftrightarrow\)X=1 (thoả mãn)

chọn A

Lời giải:
Áp dụng BĐT AM-GM:

$(x+2z)(y+2z)\leq \left(\frac{x+2z+y+2z}{2}\right)^2=\left(\frac{x+y+4z}{2}\right)^2$
$\Rightarrow \sqrt{(x+2z)(y+2z)}\leq \frac{x+y+4z}{2}$

$\Rightarrow (y+z)\sqrt{(x+2z)(y+2z)}\leq \frac{(x+y)(x+y+4z)}{2}=\frac{(x+y)^2+4zx+4zy}{2}\leq \frac{2(x^2+y^2)+2(z^2+x^2)+2(z^2+y^2)}{2}=2(x^2+y^2+z^2)$

$\Rightarrow \frac{4}{(x+y)\sqrt{(x+2z)(y+2z)}}\geq \frac{2}{x^2+y^2+z^2}$

Tương tự:

$\frac{5}{(y+z)\sqrt{(y+2x)(z+2x)}}\geq \frac{5}{2(x^2+y^2+z^2)}$

Do đó:

$P\leq \frac{4}{\sqrt{x^2+y^2+z^2+4}}-\frac{9}{2(x^2+y^2+z^2)}$

Đặt $\sqrt{x^2+y^2+z^2+4}=a$. ĐK: $a>2$ do $x,y,z$ không thể đồng thời bằng $0$

$P\leq \underbrace{\frac{4}{a}-\frac{9}{2(a^2-4)}}_{f(a)}$

$f'(a)=\frac{-4}{a^2}+\frac{9}{(a^2-4)^2}=0\Leftrightarrow a=4$

Lập bảng biến thiên suy ra:

$f(a)_{\max}=f(4)=\frac{5}{8}$

$\Rightarrow P\leq f(a)\leq \frac{5}{8}$

Vậy $P_{\max}=\frac{5}{8}$

Giá trị này đạt tại $x=y=z=2$

copy link rồi vô nhé

Đặt \(3^x=t>0\)

\(\Rightarrow\left(x+4\right)t^2-\left(x+5\right)t+1=0\)

\(\Leftrightarrow\left(x+4\right)t^2-\left(x+4\right)t-t+1=0\)

\(\Leftrightarrow\left(x+4\right)t\left(t-1\right)-\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left[\left(x+4\right)t-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\\left(x+4\right)t=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3^x=1\Leftrightarrow x=0\\\left(x+4\right)3^x=1\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow x+4=3^{-x}\)

Nhận thấy \(x=-1\) là 1 nghiệm

Ta có \(VT=x+4\) là hàm đồng biến trên R

\(VP=3^{-x}\) là hàm nghịch biến trên R

\(\Rightarrow x=-1\) là nghiệm duy nhất của (1)

Vậy \(x=\left\{0;-1\right\}\)

\(\Leftrightarrow2^{2\left(x+3\right)}+16.2^{x+3}-17=0\)

Đặt \(2^{x+3}=t>0\)

\(\Rightarrow t^2+16t-17=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-17< 0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2^{x+3}=1\)

\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

\(2\log_3\frac{x+y}{x^2+y^2+xy+2}=x^2+y^2+xy-3\left(x+y\right)\)

\(\Leftrightarrow2\log_33\left(x+y\right)+3\left(x+y\right)=2\log_3\left(x^2+y^2+xy+2\right)+\left(x^2+y^2+xy+2\right)\)

Xét hàm \(f\left(t\right)=2log_3t+t\) với \(t>0\)

\(f'\left(t\right)=\frac{2}{t.ln3}+1>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow3\left(x+y\right)=x^2+y^2+xy+2\)

\(\Rightarrow3\left(x+y\right)=\left(x+y\right)^2+2-xy\)

\(\Rightarrow-xy=3\left(x+y\right)-\left(x+y\right)^2-2\)

\(3x+2y+1=x\left(y+1\right)-xy+2\left(x+y\right)+1\)

\(\Rightarrow3x+2y+1\le\frac{1}{4}\left(x+y+1\right)^2-xy+2\left(x+y\right)+1\)

\(\Rightarrow3x+2y+1\le\frac{1}{4}\left(x+y+1\right)^2+5\left(x+y\right)-\left(x+y\right)^2-1\)

\(\Rightarrow3x+3y+1\le-\frac{3}{4}\left(x+y\right)^2+\frac{11}{2}\left(x+y\right)-\frac{3}{4}\)

Đặt \(x+y=t\Rightarrow P\le\frac{-3t^2+22t-3}{4\left(t+6\right)}\)

\(P\le\frac{-3t^2+22t-3}{4\left(t+6\right)}-1+1=\frac{-3\left(t-3\right)^2}{4\left(t+6\right)}+1\le1\)

\(P_{max}=1\) khi \(t=3\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(log_5\frac{4\sqrt{2}}{\sqrt{15}}=log_5\left(4\sqrt{2}\right)-log_5\left(\sqrt{15}\right)\)

\(=log_5\left(2^{\frac{5}{2}}\right)-log_5\left(\sqrt{3}\right)-log_5\left(\sqrt{5}\right)\)

\(=\frac{5}{2}log_52-\frac{1}{2}log_53-\frac{1}{2}\)

\(=\frac{5a-b-1}{2}\)