Lời giải:

$G=1-\sqrt{(3x-1)^2}+(3x-1)^2=1-|3x-1|+|3x-1|^2$

Đặt $|3x-1|=a$ với $a\geq 0$

Ta cần tìm GTNN của $G=1-a+a^2$

Có: $G=(a-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$ với mọi $a\geq 0$

Do đó gtnn của $G$ là $\frac{3}{4}$

Lời giải:

Ta có:

\(A=1-\sqrt{1-6x+9x^2}+(3x-1)^2=1-\sqrt{(3x-1)^2}+(3x-1)^2\)

\(=1-|3x-1|+|3x-1|^2=1-t+t^2\) (đặt \(t=|3x-1|, t\geq 0)\)

\(=(t-\frac{1}{2})^2+\frac{3}{4}\)

Ta thấy \((t-\frac{1}{2})^2\geq 0, \forall t\geq 0\)

\(\Rightarrow A=(t-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}\)

Vậy $A$ đạt min bằng $\frac{3}{4}$. Giá trị này đạt được tại $t=\frac{1}{2}\Leftrightarrow |3x-1|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} 3x-1=\frac{1}{2}\\ 3x-1=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{1}{6}\end{matrix}\right.\)

Bạn chú ý lần sau không đăng 1 bài nhiều lần tránh làm loãng box toán.

\(A=\sqrt{\left(x+1\right)^2}+\sqrt{\left(3x-1\right)^2}=\left|x+1\right|+\left|3x-1\right|\)

Với \(x\le-1:A=-x-1-3x+1=-4x\)

Để A nhỏ nhất thì x lớn nhất => x = -1 => A = 4

Với -1 < x <= 1/3: \(A=x+1-3x+1=2-2x\)

Để A nhỏ nhất thì x lớn nhất => x = 1/3 => A = 4/3

Với x > 1/3: \(A=x+1+3x-1=4x\)

Do x > 1/3 => A > 4/3

=> A min = 4/3 <=> x = 1/3

\(B=3\left(x^2-2x+\frac{1}{3}\right)=3\left[\left(x^2-2x+1\right)-\frac{2}{3}\right]=3\left(x-1\right)^2-2\)

=> Vì 3(x-1)^2 >= 0 => B >= -2

B min = -2 <=> 3(x-1)^2 = 0 <=> x = 1

\(C=2\left(x-\frac{3}{2}\sqrt{x}\right)=2\left[\left(x-2.\frac{3}{4}\sqrt{x}+\frac{9}{16}\right)-\frac{9}{16}\right]=2\left(\sqrt{x}-\frac{3}{4}\right)^2-\frac{9}{8}\)

=> C >= -9/8

C min = -9/8 <=> căn x = 3/4 => x = 9/16

\(A=1-\sqrt{1-6x+9x^2}+\left(3x-1\right)^2\)

\(A=1-\sqrt{\left(3x-1\right)^2}+\left(3x-1\right)^2\)

\(A=1-\left(3x-1\right)+\left(3x-1\right)^2\)

\(A=1-3x+1+9x^2-6x+1\)

\(A=9x^2-9x+3\)

\(A=\left(3x\right)^2-2.3x.\frac{9}{6}+\frac{81}{36}-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{3}{4}\ge0\forall x\)

Dấu = xảy ra khi:

\(3x-\frac{9}{6}=0\Leftrightarrow3x=\frac{9}{6}\Leftrightarrow x=0,5\)

Vậy A_min = -3/4 tại x = 0,5

A=1-\(\sqrt{\left(3x-1\right)^2}\)+(3x-1)^2

A=1-/3x-1/+(3x-1)^2

đặt t=/3x-1/ với t>=0

khi đó A=t^2-t+1

A=t^2-t+1/4+3/4

A=(t-1/2)^2+3/4

khi đó A>=3/4

dấu bằng xảy ra khi t=1/2 hay x=1/2

Chúc bạn học tốt!

\(A=1-\sqrt{\left(3x-1\right)^2}+\left(3x-1\right)^2\)

\(A=\left(3x-1\right)^2-\left|3x-1\right|+1\)

+) Với \(x\ge\frac{1}{3}\)\(\Rightarrow\)\(A=\left(3x-1\right)^2-\left(3x-1\right)+\frac{1}{4}+\frac{3}{4}=\left(3x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{2}\) ( tm ) 

+) Với \(x< \frac{1}{3}\)\(\Rightarrow\)\(A=\left(3x-1\right)^2+\left(3x-1\right)+\frac{1}{4}+\frac{3}{4}=\left(3x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{6}\) ( tm ) 

Vậy GTNN của \(A=\frac{3}{4}\) khi \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)

x=0 ; x=2/3 - cau b 

anh giai tu giai thu

Giai giùm đi

doi ti dc ko

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

a. \(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\)      ĐK: \(x\ge0\)

<=> \(\sqrt{9.2x}+2\sqrt{4.2x}-3\sqrt{2x}=12\)

<=> \(3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

<=> \(\sqrt{2x}\left(3+4-3\right)=12\)

<=> \(4\sqrt{2x}=12\)

<=> \(\sqrt{2x}=12:4\)

<=> \(\sqrt{2x}=3\)

<=> 2x = 3²

<=> 2x = 9

<=> \(x=\dfrac{9}{2}\) (TM)

b. \(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\)          ĐK: \(x\ge-2\)

<=> \(\sqrt{9\left(x+2\right)}+2\sqrt{36\left(x+2\right)}-\sqrt{4\left(x+2\right)}=26\)

<=> \(3\sqrt{x+2}+72\sqrt{x+2}-2\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}\left(3+72-2\right)=26\)

<=> \(73\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}=\dfrac{26}{73}\)

<=> x + 2 = \(\left(\dfrac{26}{73}\right)^2\)

<=> x + 2 = \(\dfrac{676}{5329}\)

<=> \(x=\dfrac{676}{5329}-2\)

<=> \(x=-1,873146932\) (TM)

c. \(\sqrt{\left(x-2\right)^2}=10\)

<=> \(\left|x-2\right|=10\)

<=> \(\left[{}\begin{matrix}x-2=10\left(x\ge2\right)\\x-2=-10\left(x< 2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=12\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)

d. \(\sqrt{9x^2-6x+1}=15\)

<=> \(\sqrt{\left(3x-1\right)^2}=15\)

<=> \(\left|3x-1\right|=15\)

<=> \(\left[{}\begin{matrix}3x-1=15\left(x\ge\dfrac{16}{3}\right)\\3x-1=-15\left(x< \dfrac{16}{3}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{16}{3}\left(TM\right)\\x=\dfrac{-14}{3}\left(TM\right)\end{matrix}\right.\)

e. \(\sqrt{3x+4}=3x-8\)        ĐK: \(x\ge\dfrac{-4}{3}\)

<=> 3x + 4 = (3x - 8)²

<=> 3x + 4 = 9x² - 48x + 64

<=> 9x² - 3x - 48x + 64 - 4 = 0

<=> 9x² - 51x + 60 = 0

<=> 9x² - 36x - 15x + 60 = 0

<=> 9x(x - 4) - 15(x - 4) = 0

<=> (9x - 15)(x - 4) = 0

<=> \(\left[{}\begin{matrix}9x-15=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{15}{9}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)