x= -1 nha bạn

\(\sqrt{x^2-8x+16}=5\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=5\)
\(\Leftrightarrow\left|x-4\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=5\\x-4=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\end{cases}}\)

Bài 1:

a,\(\left(\frac{1}{2}x+4\right)^2=\frac{1}{4}x^2+4x+16\)

b,(7x-5y)²=49x²-70xy+25y²

c,(6x²+y²)(y²-6x²)=(y²+6x²)(y²-6x²)=y⁴+36x⁴

Bài 2 :

a,(5x+1)³=125x³+75x²+15x+1

b,(x-2y)³=x³-6x²y+12xy²-8y³

c,(4x+5)(16x²-20x+25)=64x³+125

d,(6x-\(\frac{1}{3}\))(36x²+2x+\(\frac{1}{9}\))=216x³-\(\frac{1}{27}\)

Bài 3 :

a,(2x+3)²+(2x-3)²-2(4x²-9)

=4x²+12x+9+4x²-12x+9-8x²+18

=36

b,(x+2)³+(x-2)³+x³-3x(x+2)(x-2)

= x³+6x²+12x+8+x³-6x²+12x-8+x³-3x³+12x

=36x

Bài 4 : 

A=(3x+2)²+(2x-7)²-2(3x+2)(2x-7)

A=9x²+12x+4+4x²-28x+49-12x²+42x-8x+28

A=x²+18x+81

A=(x+9)²

Thay x=-19 vào biểu thức

=> A=(-19+9)²

     A=(-10)²=100

Bài 5 :

B=(3x-1)²-(x+7)²-2(2x-5)(2x+5)

B= 9x²-6x+1-x²+14x-49-8x²+50

B=8x+2

B=2(4x+1)

Thay x=\(\frac{1}{5}\)vào biểu thức

=> B=2(4.\(\frac{1}{5}\)+1)

      B=2.\(\frac{9}{5}\)=\(\frac{18}{5}\)

\(A=\sqrt{64a^2}\cdot2a=\sqrt{\left(8a\right)^2}\cdot2a=\left|8a\right|\cdot2a\)

Với a < 0 A = 8a.(-2a) = -16a²

Với a ≥ 0 A = 8a.2a = 16a²

\(B=3\sqrt{9a^6}-6a^3=3\sqrt{\left(3a^3\right)^2}-6a^3=9\left|a^3\right|-6a^3\)

a) ( x² - 3x )( x² + 7x + 10 ) = 216

<=> x( x - 3 )( x + 2 )( x + 5 ) - 216 = 0

<=> [ x( x + 2 ) ][ ( x - 3 )( x + 5 ) ] - 216 = 0

<=> ( x² + 2x )( x² + 2x - 15 ) - 216 = 0 (1)

Đặt a = x² + 2x

(1) trở thành a( a - 15 ) - 216 = 0 <=> a² - 15a - 216 = 0 <=> ( a - 24 )( a + 9 ) = 0 <=> a = 24 hoặc a = -9

=> \(\orbr{\begin{cases}x^2+2x=24\\x^2+2x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2x-24=0\\x^2+2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-4\right)\left(x+6\right)=0\\\left(x+1\right)^2+8>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

Vậy S = { 4 ; -6 }

b) ( 2x² - 7x + 3 )( 2x² + x - 3 ) + 9 = 0

<=> ( x - 3 )( 2x - 1 )( x - 1 )( 2x + 3 ) + 9 = 0

<=> [ ( x - 3 )( 2x + 3 ) ][ ( 2x - 1 )( x - 1 ) ] + 9 = 0

<=> ( 2x² - 3x - 9 )( 2x² - 3x + 1 ) + 9 = 0

<=> ( 2x² - 3x - 4 - 5 )( 2x² - 3x - 4 + 5 ) + 9 = 0

<=> ( 2x² - 3x - 4 )² - 16 = 0

<=> x( 2x - 3 )( 2x² - 3x - 8 ) = 0

<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x² - 3x - 8 = 0

<=> x = 0 hoặc x = 3/2 hoặc x = \(\frac{3\pm\sqrt{73}}{4}\)

Vậy S = { 0 ; 3/2 ; \(\frac{3\pm\sqrt{73}}{4}\)}

phải là (\(\frac{3}{\sqrt{1+a}}\)+\(\sqrt{1+a}\)): (\(\frac{3}{\sqrt{1-a^2}}\)+1) nha! LÀ dấu CỘNG , KO PHẢI TRỪ:>

F = [3/(√1 + a) + (√1 - a)] : [3/(√1 - a^2) + 1] .

=[3+√1-a^2)/√1+a]:[(3+√1-a^2/√1-a^2] 

=(3+√1-a^2/√1+a].[√1-a^2/(3+√1-a^2]

=√1-a.√1+a/√1+a=√1-a

thay a=√3/(2+√3) vào F ta được

√[1-(√3/(2+√3)]=√2/(2+√3)

=√2/(2+√3)=(√4-2√3)/4-3=√(√3-1)^2=|√3-1}

=√3-1

Ta có: \(\sqrt{x-7}\le\frac{x-7+1}{2}=\frac{x-6}{2}\)(bđt cosi)

 \(\sqrt{9-x}\le\frac{9-x+1}{2}=\frac{10-x}{2}\)

=> \(VT=\sqrt{x-7}+\sqrt{9-x}\le\frac{x-6}{2}+\frac{10-x}{2}=\frac{x-6+10-x}{2}=2\)

\(VP=x^2-16x+66=\left(x-8\right)^2+2\ge2\)

=> \(VT=VP\Leftrightarrow\hept{\begin{cases}x-7=1\\9-x=1\\x-8=0\end{cases}}\) <=> x = 8

Vậy S = {8}

\(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\left(7\le x\le9\right)\)

Đặt \(A=\sqrt{x-7}+\sqrt{9-x}\)

\(\Rightarrow A^2=\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\)

Áp dụng bất đảng thức Bunhiacopxki ta có:

\(\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\le\left(x-7+9-x\right)\left(1+1\right)=4\)

=> \(A\le2\)

Ta có: \(x^2-16x+66=\left(x-8\right)^2+2\ge2\)

Dấu = xảy ra

\(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-7}}{1}=\frac{\sqrt{9-x}}{1}\\x-8=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=\sqrt{9-x}\\x=8\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-7=9-x\\x=8\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=8\\x=8\end{cases}\left(tm\right)}\)

Vậy x = 8

Đk: \(\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\)

Ta có: \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

VT = \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}\ge0\) => >VP = 2x + 2 \(\ge\)0 => x \(\ge\)-1

mà \(\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\) => \(x\ge1\)

<=> \(\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\)

<=> \(\sqrt{x+1}.\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)

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

<=> \(2x+6+x-1+2\sqrt{2\left(x-1\right)\left(x+3\right)}=4x+4\)

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

<=> \(\sqrt{x-1}.\left(2\sqrt{2\left(x+3\right)}-\sqrt{x-1}\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\2\sqrt{2\left(x+3\right)}=\sqrt{x-1}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\8x+24=x-1\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{25}{7}\left(ktm\right)\end{cases}}\)

Vậy S = {1}

\(\sqrt{3+x}+\sqrt{6-x}-\sqrt{\left(3+x\right)\left(6-x\right)}=3\)(ĐK: \(-3\le x\le6\))

Đặt \(t=\sqrt{3+x}+\sqrt{6-x}\ge0\)

\(\Leftrightarrow t^2=3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}\)

\(\Leftrightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=\frac{t^2-9}{2}\)

Phương trình ban đầu tương đương với: 

\(t-\frac{t^2-9}{2}=3\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\left(tm\right)\\t=-1\left(l\right)\end{cases}}\)

Với \(t=3\):

\(\sqrt{3+x}+\sqrt{6-x}=3\)

\(\Leftrightarrow9+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-3\left(tm\right)\\x=6\left(tm\right)\end{cases}}\)

\(\sqrt{2x+1}-\sqrt{2-x}=x+3\)(ĐK: \(-\frac{1}{2}\le x\le2\))

Ta có: \(\sqrt{2x+1}-\sqrt{2-x}\le\sqrt{2x+1}\le\sqrt{2.2+1}=\sqrt{5}\)(vì \(-\frac{1}{2}\le x\le2\))

\(x+3\ge-\frac{1}{2}+3=2,5\)(vì \(-\frac{1}{2}\le x\le2\))

mà \(\sqrt{5}< 2,5\)

do đó phương trình vô nghiệm.