=232/77

Mí bn ghi cả cách làm giúp mik nha

a: Ta có: \(A=x^2-7x+11\)

\(=x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{5}{4}\)

\(=\left(x-\dfrac{7}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{7}{2}\)

b: ta có: \(A=9x^2+6x+11\)

\(=9x^2+6x+1+10\)

\(=\left(3x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{3}\)

\(\frac{1}{3}x-1.2x-\frac{2}{23}.\left(\frac{-11}{12}+\frac{5}{6}-\frac{7}{8}\right)-\frac{1}{8}=-11\frac{2}{3}+12\frac{7}{8}\)

\(\Leftrightarrow\frac{1}{3}x-\frac{6}{5}x-\frac{2}{23}.\frac{-23}{24}-\frac{1}{8}=\frac{-35}{3}+\frac{103}{8}\)

\(\Leftrightarrow\frac{-13}{15}x+\frac{1}{12}-\frac{1}{8}=\frac{29}{24}\)

\(\Leftrightarrow\frac{-13}{15}x-\frac{1}{24}=\frac{29}{24}\)

\(\Leftrightarrow\frac{-13}{15}x=\frac{5}{4}\)

\(\Leftrightarrow x=\frac{-75}{52}\)

Vậy \(x=\frac{-75}{52}\)

1/3x-1/2x-2/23.(-11/12+5/6-7/8)-1/8=-31/3+103/8

1/3x-1,2x-2/23.(-11/12+5/6-7/8)-1/8=61/24

1/3x-1,2x-2/23.-23/24-1/8=61/24

1/3x-1,2x-1/12-1/8=61/24

1/3x-1,2x+1/24=61/24

x.(1/3-1,2)+1/24=61/24

x.-13/15+1/24=61/24

x.-33/40=61/24

x=61/24:-33/40

x=61/24×-40/33

x=-305/99

vậy x=-305/99

a: A=-(x-7)^2-888<=-888

Dấu = xảy ra khi x=7

b: \(B=\left|2x-1\right|+\left|y-5\right|+\dfrac{8}{3}>=\dfrac{8}{3}\)

Dấu = xảy ra khi x=1/2 và y=5

c: \(C=\left(x+3\right)^2+\left|2y-5\right|-232>=-232\)

Dấu = xảy ra khi x=-3 và y=5/2

giê ơt nha bn

a, 3x - x - 3 = 11 

=> 2x - 3 = 11 

=> 2x = 14 

=> x = 7

b, (x-2)(x-5) = 0 

=> \(\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=5\end{cases}}\)

c, d là tương tự

bài 2 

a, (x-1)(y+3) = 5 

=> \(\hept{\begin{cases}x-1=1\\y+3=5\end{cases}}\)=> \(\hept{\begin{cases}x=2\\y=2\end{cases}}\)

=> \(\hept{\begin{cases}x-1=5\\y+3=1\end{cases}}\)=> \(\hept{\begin{cases}x=6\\y=-2\end{cases}}\)

câu sau tương tự

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

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

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

A= \(\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)

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

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

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

\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|\)

=2.

dấu = khi và chỉ khi \(\left(\sqrt{x-1}+1\right).\left(1-\sqrt{x-1}\right)=0\)

=0 nha bn