\(\left(x^2+16\right)^2-\left(16x-1\right)=0\)

\(\Leftrightarrow x^4+32x^2-16x+257=0\)

\(\Leftrightarrow x^4+24x^2+\left(8x^2-16x+8\right)+249=0\)

\(\Leftrightarrow x^4+24x^2+8\left(x-1\right)^2+249=0\)

Dễ thấy VP > 0

Vậy PT vô nghiệm

Rảnh rỗi thật sự .-.

bài 11

a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)

b)

\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)

c)

\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

bài 12

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x=26\\ x=-2\)

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

a) (−11)² − 15(x − 2) = 134 − 16x

121 - 15x + 30 = 134 - 16x

16x - 15x = 134 - 121 - 30

x = -17

b) (4x + 1)(x² − 16)=0

(4x + 1)(x - 4)(x + 4) = 0

\(\left[\begin{matrix}4x+1=0\\x-4=0\\x+4=0\end{matrix}\right.\)

\(\left[\begin{matrix}4x=-1\\x=4\\x=-4\end{matrix}\right.\)

\(\left[\begin{matrix}x=-\frac{1}{4}\\x=4\\x=-4\end{matrix}\right.\)

c) − 2(x − 3) + (− 2)² = 4 − 3x

3x + 4 - 2x + 6 = 4

x = 4 - 4 - 6

x = - 6

Nguyễn Huy TúPhương An

1) \(\sqrt[]{9\left(x-1\right)}=21\)

\(\Leftrightarrow9\left(x-1\right)=21^2\)

\(\Leftrightarrow9\left(x-1\right)=441\)

\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)

2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)

\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)

\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)

mà \(\sqrt[]{1-x}\ge0\)

\(\Leftrightarrow pt.vô.nghiệm\)

3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)

\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)

\(\Leftrightarrow2x=50\Leftrightarrow x=25\)

1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))

\(\Leftrightarrow3\sqrt{x-1}=21\)

\(\Leftrightarrow\sqrt{x-1}=7\)

\(\Leftrightarrow x-1=49\)

\(\Leftrightarrow x=49+1\)

\(\Leftrightarrow x=50\left(tm\right)\)

2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))

\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)

\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)

\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý) 

Phương trình vô nghiệm

3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\)) 

\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)

\(\Leftrightarrow2x=50\)

\(\Leftrightarrow x=\dfrac{50}{2}\)

\(\Leftrightarrow x=25\left(tm\right)\)

4) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

5) \(\sqrt{\left(x-3\right)^2}=3-x\)

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

\(\Leftrightarrow x-3=3-x\)

\(\Leftrightarrow x+x=3+3\)

\(\Leftrightarrow x=\dfrac{6}{2}\)

\(\Leftrightarrow x=3\)

1) => 9(x-1)=\(21^2\)

=> 9x-9=441

=> 9x=450

=> x=50

2)=>\(\sqrt{1-x}\) + \(\sqrt{4\left(1-x\right)}\)-\(\dfrac{1}{3}\sqrt{16\left(1-x\right)}\)+5=0

=>\(\sqrt{1-x}\)\(\left(1+2-\dfrac{1}{3}.4\right)\)+5=0

=>\(\dfrac{5}{3}\sqrt{1-x}\) +5=0

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

Phuong trinh vo nghiem

(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)

cái này đâu ra z ???

nguyen van tuan: hì, xin lỗi, làm hơi tắt ^^!

\(\left(1\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}=\left(x+1\right)\left(x-\dfrac{23}{8}\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}-\left(x+1\right)\left(x-\dfrac{23}{8}\right)=0\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)

a/\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)

\(\Leftrightarrow4x^4+16x^3+23x^2+14x-15=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)\left(x^2+2x+3\right)=0\)

Tới đây thì đơn giản rồi b tự làm nhé

b/ \(3x^4-13x^3+16x^2-13x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x-1\right)\left(x^2-x+1\right)=0\)

Tới đây thì bạn làm tiếp nhé

c/ \(\left(x+3\right)^4+\left(x+5\right)^4=16\)

\(\Leftrightarrow2x^4+32x^3+204x^2+608x+690=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(x^2+8x+23\right)=0\)

Bạn làm tiếp nhé

a, \(x^2-3x+2=0\\ < =>x^2-x-2x+2=0\\ < =>\left(x^2-x\right)-\left(2x-2\right)=0\\ < =>x\left(x-1\right)-2\left(x-1\right)=0\\ < =>\left(x-2\right)\left(x-1\right)=0\\ < =>\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

b) x³ + x² - 36 = 0

=> x².(x + 1) = 36

Vì x² \(\ge\) 0 => (x + 1) \(\ge\) 0 (1)

Mặt khác: x² là số chính phương nên những tích ko có số chính phương sẽ bị loại (2)

Từ điều kiện (1) và (2),ta có các TH sau:

TH₁ : x².(x + 1) = 1.36

=> \(\left\{{}\begin{matrix}x^2=1\\x+1=36\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1;-1\\x=35\end{matrix}\right.\) => Loại

TH_2: x².(x+1) = 36.1

=> \(\left\{{}\begin{matrix}x^2=36\\x+1=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=6;-6\\x=0\end{matrix}\right.\) => Loại

TH₃: x².(x + 1) = 4.9

=> \(\left\{{}\begin{matrix}x^2=4\\x+1=9\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=2;-2\\x=8\end{matrix}\right.\) => Loại

TH₄ : x².(x + 1) = 9.4

=> \(\left\{{}\begin{matrix}x^2=9\\x+1=4\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=3;-3\\x=3\end{matrix}\right.\) => x = 3

Vậy x = 3

P/s: Đây là cách của mk. Bạn cx có thể í luận thêm để loại bỏ thêm 1 số TH nhé!!!

b)\(x^3+x^2-36=0\)

<=>\(\left(x-3\right)\left(x^2+4x+12\right)=0\)(1)

Vì \(x^2+4x+12=\left(x+2\right)^2+8>0\) với mọi x nên:

(1)<=>x-3=0

<=>x=3

Vậy x=3.