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TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
Câu 1 :
a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)
b.
|-2x|=5x+14
Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14
<=> - 2x = 5x + 14
<=> - 2x - 5x = 14
<=> - 7x = 14
<=> x = - 2 (thoã mãn)
Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.
Ta có pt: 2x = 5x + 14
<=> - 3x = 14
<=> x = \(-\dfrac{14}{3}\)
Vậy pt có nghiệm x = - 2
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)
\(a,4x-5=23\)
\(\Leftrightarrow4x=23+5\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(b,\left|-2x\right|=5x+14\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)
\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)
\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
1/ \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\) (1)
Điều kiện: \(\left\{{}\begin{matrix}x-1\ne0\\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)
(1) \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\\ \Leftrightarrow\left(16+21+12-5\right)x=-5-28\\ \Leftrightarrow44x=-33\\ \Leftrightarrow x=-\dfrac{3}{4}\) (Thỏa mãn)
Vậy \(x=-\dfrac{3}{4}\).
2/ \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\) (2)
Điều kiện: \(x\ne\pm1\)
(2)\(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow x\left(x+1\right)-2x=0\\ \Leftrightarrow x^2+x-2x=0\\ \Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
So sánh với điều kiện \(\Rightarrow x=0\) là nghiệm của PT.
3/ \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\) (3)
Điều kiện: \(x\ne\pm3\)
(3)\(\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\\ \Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ \Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\\ \Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0\) (Vô nghiệm do \(x^2+x+8>0\qquad\forall x\)).
Vậy PT vô nghiệm.
4/ \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\) (4)
Điều kiện: \(x\ne\pm1\)
(4)\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1\) (loại)
Vậy PT vô nghiệm.
5/ \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) (5)
Điều kiện: \(x\ne0\)
(5)\(\Leftrightarrow x+\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^2-2\)
Đặt \(t=x+\dfrac{1}{x}\), ta có: \(t=t^2-2\\ \Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-1\end{matrix}\right.\)
Với \(t=2\) ta có: \(x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\) (thỏa mãn)
Với \(t=-1\) ta có: \(x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0\) (vô nghiệm).
Vậy \(x=1\) là nghiệm PT.
6/ \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\) (6)
Điều kiện: \(x\ne-1\)
(6)\(\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.\)
\(x-1=0\Leftrightarrow x=1\) (Thỏa mãn)
\(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\\ \Leftrightarrow x^2-x+3=0\) (vô nghiệm).
Vậy \(x=1\) là nghiệm PT.
1) ĐKXĐ: \(x\notin\left\{1;-\dfrac{4}{3}\right\}\)
Ta có: \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0\)
\(\Leftrightarrow20x+33=0\)
\(\Leftrightarrow20x=-33\)
\(\Leftrightarrow x=-\dfrac{33}{20}\)(nhận)
Vậy: \(S=\left\{-\dfrac{33}{20}\right\}\)
2) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
Suy ra: \(x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
3) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\)
\(\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(-x-3-14=x^2-9\)
\(\Leftrightarrow x^2-9=-x-17\)
\(\Leftrightarrow x^2-9+x+17=0\)
\(\Leftrightarrow x^2+x+8=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0\)(vô lý)
Vậy: \(S=\varnothing\)
4) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
5) ĐKXĐ: \(x\ne0\)
Ta có: \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{x^4+1}{x^2}\)
\(\Leftrightarrow x^2\left(x^2+1\right)=x\left(x^4+1\right)\)
\(\Leftrightarrow x^4+x^2=x^5+x\)
\(\Leftrightarrow x^5+x-x^4-x^2=0\)
\(\Leftrightarrow x\left(x^4-x^3-x+1\right)=0\)
\(\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\)
nên \(x\cdot\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x-1=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy: S={1}
6) ĐKXĐ: \(x\in R\)
Ta có: \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+4\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1-x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2+x-3\right)=0\)
\(\Leftrightarrow-\left(x-1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\)
nên x-1=0
hay x=1(nhận)
Vậy: S={1}
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý
b.
Đặt \(x+3=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)
\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)
\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)
Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.
2:
a: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{24}{9}=\dfrac{8}{3}\)
=>x=16/3; y=8; z=32/3
A=3x+2y-6z
=3*16/3+2*8-6*32/3
=16+16-64
=-32
b: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y+z}{5-6+7}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
=>x=5căn 2; y=6căn 2; y=7căn 2
B=xy-yz
=y(x-z)
=6căn 2(5căn 2-7căn 2)
=-6căn 2*2căn 2
=-24
b
C