$\sqrt[3]{33x-5}=8x^3-36x^2+53x-25$

$PT\iff \sqrt[3]{33x-5}=(2x-3{)}^3-(x-2)$

Đặt $y=\sqrt[3]{33x-5}\Rightarrow \{\begin{array}{c}{y}^3=3x-5=(2x-3)+(x-2)\\ y=(2x-3{)}^3-(x-2)\end{array}$

$\Rightarrow y^3+y=(2x-3{)}^3+(2x-3)$ (1)
Xét hàm: $f\left(t\right)=t^3+t$
có $f^{{}^{}}\left(t\right)=3t^2+1>0$ nên là hàm đồng biến (2)
Từ (1) và (2) suy ra $y=2x-3$
Đến đây thay vào , giải PT bậc 3

Chỉ bk lm trừ, ko bk lm cộng

đặt \(\sqrt{3x+1}=a\) 

=> pt <=> 4x^2 +a +6=a^2 +12x

chuyển hết nt sang vế phải để vt =0 ptđttnt có ntc=a+2x-3

câu 2 đặt \(\sqrt[3]{3x-5}=2y-3\) rồi làm tt như bài trên lớp

sau khi chuyển  cậu có pt a62-4x^2-a+12x-6=0

=> a^2+2ax-3a-2ax-4x^2+6x+2a+4x-6=0

<=> (a+2x-3)(a-2x+2)=0

c2 đăt...

=>3x-5=(2y-3)^3 

mặt khác từ pt =>\(\sqrt[3]{3x-5}=\left(2x-3\right)^3-x+2\)

=>2y-3=(2x-3)^3-x+2

=>2y+x-5=(2x-3)^3 rồi cậu giải tt bài trên lớp

1/ Đk : \(2x^2-6x-1\ge0\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{3-\sqrt{11}}{2}\\x\ge\frac{3+\sqrt{11}}{2}\end{matrix}\right.\)

Bình phương 2 vế của phương trình, ta có :

\(4x^4+36x^2+1-24x^3-4x^2+12x-4x-5=0\)

\(\Leftrightarrow4x^4-24x^3+32x^2+8x-4=0\)

\(\left[{}\begin{matrix}x=1-\sqrt{2}\left(TM\right)\\x=2-\sqrt{3}\left(l\right)\\x=\sqrt{2}+1\left(l\right)\\x=\sqrt{3}+2\left(TM\right)\end{matrix}\right.\)

Vậy ....

\(\Leftrightarrow8x^3-36x^2+51x-22+2x-3-\sqrt[3]{3x-5}=0\)

\(\Leftrightarrow8x^3-36x^2+51x-22+\dfrac{8x^3-36x^2+51x-22}{\left(2x-3\right)^2+\left(2x-3\right)\sqrt[3]{3x-5}+\sqrt[3]{\left(3x-5\right)^2}}=0\)

\(\Leftrightarrow\left(8x^3-36x^2+51x-22\right)\left(1+\dfrac{1}{\left(2x-3\right)^2+\left(2x-3\right)\sqrt[3]{3x-5}+\sqrt[3]{\left(3x-5\right)^2}}\right)=0\)

\(\Leftrightarrow8x^3-36x^2+51x-22=0\)

\(\Leftrightarrow\left(x-2\right)\left(8x^2-20x+11\right)=0\)

\(\Leftrightarrow...\)

Cách khác: (Đưa về hàm đặc trưng)

\(PT\Leftrightarrow8x^3-36x^2+53x-25=\sqrt[3]{3x-5}\)

\(\Leftrightarrow\left(2x-3\right)^3+2x-3=3x-5+\sqrt[3]{3x-5}\). (*)

Xét hàm \(f\left(t\right)=t^3+t\). Ta thấy f(t) đồng biến trên \(\mathbb{R}\).

Do đó \(\left(\cdot\right)\Leftrightarrow2x-3=\sqrt[3]{3x-5}\)

\(\Leftrightarrow8x^3-36x^2+54x-27=3x-5\)

\(\Leftrightarrow8x^3-36x^2+51x-22=0\)

\(\Leftrightarrow\left(x-2\right)\left(8x^2-20x+11\right)=0\Leftrightarrow...\)

chịu thôi

\(2.< =>5\sqrt{x-1}-6\sqrt{x-1}-3\sqrt{x-1}=2\sqrt{2x-3}\)

\(< =>\sqrt{x-1}\left(5-6+3\right)=2\sqrt{2x-3}\)

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

\(< =>x-1=2x-3\)

\(< =>x=2\)

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)