Ta có

\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)

\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)

\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)

Thế vào ta được

\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)

\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)

ĐKXĐ: \(x\ge3\)

\(\sqrt{x-1}>\sqrt{x-2}+\sqrt{x-3}\)

\(\Leftrightarrow x-1>2x-5+2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow4-x>2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\\left(4-x\right)^2>4\left(x^2-5x+6\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\3x^2-12x+8< 0\end{matrix}\right.\)

\(\Rightarrow\dfrac{6-2\sqrt{3}}{3}< x< \dfrac{6+2\sqrt{3}}{3}\)

Kết hợp ĐKXĐ \(\Rightarrow3\le x< \dfrac{6+2\sqrt{3}}{3}\)

\(=\dfrac{5-3\sqrt{5}+10+6\sqrt{5}}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}-\dfrac{2\sqrt{10}+2}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{15+3\sqrt{5}}{5-9}-\left(2\sqrt{10}+2\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =-2\sqrt{30}-4\sqrt{5}-2\sqrt{3}-2\sqrt{2}-\dfrac{15+3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-16\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15-3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-19\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15}{4}\)

a) \(\Leftrightarrow\left(-63x^2+78x-15\right)+\left(63x^3+x-20\right)=44\)

\(\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\)

\(\Leftrightarrow79x-35=44\)

\(\Leftrightarrow79x=44+35\)

\(\Leftrightarrow79x=79\)

\(\Leftrightarrow x=1\)

b) \(\Leftrightarrow\left(x^2+3x+2\right).\left(x+5\right)-x^2.\left(x+8\right)=27\)

\(\Leftrightarrow x.\left(x^2+3x+2\right)+5.\left(x^2+3x+2\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x+10=27\)

\(\Leftrightarrow17x=17\)

\(\Leftrightarrow x=1\)

bạn chỉ cần cố gắng là làm được

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

a) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}\)<\(\dfrac{3-x}{5}-\dfrac{2x-1}{4}\)

=> 20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)

<=>40x-100-90x+30<36-12x-30x+15

<=>-50x-70<51-42x

<=>-50x+42x<51+70

<=> -8<121

<=>x>\(\dfrac{-121}{8}\)

=> S={x|x>\(\dfrac{-121}{8}\)}

b) 5x-\(\dfrac{3-2x}{2}\)>\(\dfrac{7x-5}{2}\)+x

=> 10x-(3-2x)>7x-5+2x

<=>10x-3+2x>7x-5+2x

<=>10x-3>7x-5

<=>10x-7x>-5+3

<=>3x>-2

<=>x>\(\dfrac{-2}{3}\)

=>S={x|x>\(\dfrac{-2}{3}\)}

c)\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\)

=>4(7x-2)-24x<60-3(x-2)

<=> 28x-8-24x<60-3x+6

<=>4x-8<66-3x

<=>4x+3x<66+8

<=>7x<74

<=>x<\(\dfrac{74}{7}\)

=>S={x|x<\(\dfrac{74}{7}\)}

Lời giải:

a) Vì \(6^x-2^x>0\Rightarrow x>0\)

Xét \(y=6^x-2^x-32\) có \(y'=\ln 6.6^x-\ln 2.2^x>0\forall x>0\) nên hàm $y$ đồng biến trên \(x\in(0,+\infty)\).

Khi đó phương trình \(6^x-2^x=32\) có nghiệm duy nhất $x=2$

b) Có \(5^{7^x}=7^{5^x}\Leftrightarrow \log(5^{7^x})=\log (7^{5^x})\)

\(\Leftrightarrow 7^x\log 5=5^x\log 7=7^{x\frac{\log 5}{\log 7}}\log 7\)

\(\Leftrightarrow 7^{x(1-\frac{\log 5}{\log 7})}=\frac{\log 7}{\log 5}=10^{x\log 7(1-\frac{\log 5}{\log 7})}=10^{x\log(\frac{7}{5})}\)

\(\Leftrightarrow x\log\frac{7}{5}=\log\left ( \frac{\log 7}{\log 5} \right )\)\(\Rightarrow x=\frac{\log\left ( \frac{\log 7}{\log 5} \right )}{\log\frac{7}{5}}\)

d) ĐKXĐ:...........

\(3^x+\frac{1}{3^x}=\sqrt{8-x^2}\Leftrightarrow 9^x+\frac{1}{9^x}+2=8-x^2\)

\(\Leftrightarrow 9^x+\frac{1}{9^x}+x^2=6\)

Giả sử \(x\geq 0\) . Xét hàm \(y=9^x+\frac{1}{9^x}+x^2\) có \(y'=9^x\ln 9-\frac{\ln 9}{9^x}+2x\geq 0\) nên hàm đồng biến trên \(x\in [0,+\infty)\)

Do đó PT \(9^x+\frac{1}{9^x}+x^2=6\) với $x\geq 0$ có nghiệm duy nhất \(x\approx 0,753897\)

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Vì hàm \(y\) là hàm chẵn nên $-x$ cũng là nghiệm, do đó tổng kết lại PT có nghiệm là \(x\approx \pm 0,753897\)

e) ĐK: $x>0$

\(\text{PT}\Leftrightarrow x^{\log_29}+x^{\log_23}=x^2.3^{\log_2x}\)

\(\Leftrightarrow x^{2\log_23}+x^{log_23}=x^2.x^{log_23}\Leftrightarrow x^{log_23}(x^{\log_23}+1-x^2)=0\)

\(\Leftrightarrow x^{\log_23}+1-x^2=0\) (do \(x>0\))

Dễ thấy \(x^2>x^{\log_23}\Rightarrow x>1\)

Xét hàm \(y=x^2-x^{\log_23}\Rightarrow y'=2x-\log_23x^{\log_23-1}>0\forall x>1\) nên hàm $y$ là hàm đồng biến

Do đó PT có nghiệm duy nhất $x=2$

c) Có lẽ bạn type thiếu đề